Skip to main content

Full text of "Cyclopedia of civil engineering; a general reference work on surveying, railroad engineering, structural engineering, roofs and bridges, masonry and reinforced concrete, highway construction, hydraulic engineering, irrigation, river and harbor improvement, municipal engineering, cost analysis, etc"

See other formats


( ^ 








Copyright, 1907, by Underwood <t Underwood, Net 


Civil 6- Mechanical Engineer. 




Civil Engineering 

A General Reference Work 






Editor-in- Chief 


Assisted by a Corps of 


Illustrated with over Three Thousand Engravings 








Entered at Stationers' Hall, London 
All Rights Reserved, 





Editor-in- Chief 

Dean, CoJJege of Engineering, University of Wisconsin 

Authors and Collaborators 


Consulting Civil Engineer 

American Society of Civil Engineers 

Author of "Railroad Construction, ' "Economics of Railroad Construction," etc. 


Assistant Professor of Structural Engineering, University of Illinois 
American Society of Civil Engineers 
American Society for Testing Materials 


Consulting Engineer 
American Society of Civil Engineers 
Managing Editor "Engineering-Contracting" 

Author of "Handbook of Cost Data for Contractors and Engineers," "Earthwork 
and its Cost," "Rock Excavation Methods and Cost" 


Adjunct Professor of Civil Engineering, Columbia University, N. Y. 


Professor of Mechanics, University of Wisconsin 

Joint Author of "Principles of Reinforced Concrete Construction" 


Civil Engineer 

Designer of Reinforced Concrete 


Civil Engineer 

Author of "Highway Construction," "Materials and Workmanship" 


Authors and Collaborators Continued 


Dean of the College of Engineering, and Professor of Engineering, University of 


American Society of Civil Engineers 
Joint Author of "Principles of Reinforced Concrete Construction," "Public Water 

Supplies," etc. 


Instructor in Civil Engineering, American School of Correspondence 

Formerly with Engineering Department, Atchison, Topeka & Santa Fe Railroad 


Head of Department of Civil Engineering, Armour Institute of Technology 


Instructor in Mechanical Engineering, American School of Correspondence 


Instructor in Civil Engineering. Columbia University, N. Y. 
Author of "Highway Engineering." 


Instructor in Mechanical Drawing, Massachusetts Institute of Technology 


Head of Instruction Department, American School of Correspondence 
American Society of Mechanical Engineers 
Western Society of Engineers 


A rchitectural Engineer 

American Society of Civil Engineers 


Instructor in Electrical Engineering, American School of Correspondence 
American Institute of Electrical Engineers 


Dean of Division of Engineering and Professor of Civil Engineering, Iowa State 


American Society of Civil Engineers 
Western Society of Civil Engineers 

Authors and Collaborators Continued 


Civil and Sanitary Engineer 

Chief Sanitary Inspector, City of Chicago 

American Society of Civil Engineers 


Secretary, American School of Correspondence 


Massachusetts Institute of Technology 
Ecole des Beaux Arts, Paris 


Consulting Engineer 

American Society of Civil Engineers 

Chief Engineer, Construction Service Co. 


Textbook Department, American School of Correspondenc 
Formerly Instructor, Cornell University 
Royal Astronomical Society of Canada 



Sanitary Engineer 

National Association of Master Plumbers 

United Association of Journeyman Plumbers 


R. T. MILLER, Jr., A. M., LL. B. 

President American School of Correspondence 


Head of Technical Publication Department, Westinghouse Electric & Manufac- 
turing Co. 

Formerly Technical Editor, "Street- Rail way Review" 
Formerly Editor "The Technical World Magazine" 


Massachusetts Institute of Technology 

British Society of Chemical Industry, American Chemical Society, etc. 

HARRIS C. TROW, S. B., Managing Editor 

Editor of Textbook Department, American School of Correspondence 
American Institute of Electrical Engineers 

Authorities Consulted 

THE editors have freely consulted the standard technical literature of 
America and Europe in the preparation of these volumes. They desire 
to express their indebtedness, particularly, to the following eminent 
authorities, whose well-known treatises should be in the library of everyone 
interested in Civil Engineering. 

Grateful acknowledgment is here made also for the invaluable co-opera- 
tion of the foremost Civil, Structural, Railroad, Hydraulic, and Sanitary 
Engineers in making these volumes thoroughly representative of the very 
best and latest practice in every branch of the broad field of Civil Engineer- 
ing; also for the valuable drawings and data, illustrations, suggestions, 
criticisms, and other courtesies. 


Dean of the School of Applied Science and Professor of Civil Engineering in the State 

University of Iowa; American Society of Civil Engineers. 
Author of "A Textbook of Plane Surveying," "The Elements of Railroad Engineering." 



Hydraulic Engineer and Water-Power Expert; American Society of Civil Engineers. 
Author of " Water Power, the Development and Application of the Energy of Flowing 


Dean of the College of Engineering and Professor of Engineering, University of Wisconsin. 
Joint Author of "Public Water Supplies," "Theory and Practice of Modern Framed 
Structures," " Principles of Reinforced Concrete Construction." 


H. N. OGDEN, C. E. 

Assistant Professor of Civil Engineering, Cornell University. 
Author of "Sewer Design." 



Professor of Civil Engineering in the Western University of Pennsylvania. 
Author of "A Treatise on Plane Surveying." 


Editor of Engineering-Contracting; American Society of Civil Engineers; Late Chief 

Engineer, Washington State Railroad Commission. 
Author of " Handbook of Cost Data for Contractors and Engineers." 



Late Professor of Civil Engineering, University of Michigan. 

Author of "Trusses and Arches, Graphic Method," "Structural Mechanics." 

Authorities Consulted Continued 


Editor of Municipal Journal and Engineer; Formerly Professor of Municipal Engineer- 
ing, Lafayette College. 
Author of " Water Supply Engineering," "Sewerage." 


Emeritus Professor of Civil Engineering and Surveying. University College, London; 
Institution of Civil Engineers. 

Author of "Rivers and Canals," "Harbors and Docks," "Achievements in Engineer- 
ing," "Civil Engineering as Applied in Construction." 


Professor of Civil Engineering, Syracuse University. 
Author of " Plane Surveying." 


Consulting Engineer; Associate Editor of The Engineering Record; Non-Resident Lec- 
turer on Field Engineering in Cornell University. 
Author of " Types and Details of Bridge Construction." 


Member of the Institution of Civil Engineers. 
Author of " Irrigation, Its Principles and Practice." 



American Society of Civil Engineers. 

Joint Author of "A Treatise on Concrete, Plain and Reinforced." 



American Society of Civil Engineers. 

Author of "Architectural Engineering," "Fireproofing of Steel Buildings." 


Civil Engineer. 

Author of "Highway Construction," "Inspection of Materials and Workmanship Em- 
ployed in Construction." 


Inspector of Geodetic Work and Chief of Computing Division, Coast and Geodetic Survey; 

American Society of Civil Engineers. 
Author of "A Textbook of Geodetic Astronomy." 


Consulting Civil Engineer; American Society of Civil Engineers. 

Author of "Railroad Construction in Theory and Practice," "Economics of Railroad 
Construction, "etc. 

Authorities Consulted Continued 


Professor of Mechanics, University of Wisconsin. 
Joint Author of " Principles of Reinforced Concrete Construction." 


Geographer and Former Irrigation Engineer, United States Geological Survey; American 

Society of Civil Engineers. 
Author of " Topographic Surveying," " Irrigation Engineering," etc. 


Professor of Civil Engineering, Lehigh University. 

Author of " The Elements of Precise Surveying and Geodesy," "A Treatise on Hydraul- 
ics," "Mechanics of Materials," " Retaining Walls and Masonry Dams, " "Introduction 
to Geodetic Surveying," "A Textbook on Roofs and Bridges," "A Handbook for 
Surveyors,'' etc. 



American Society of Civil Engineers; Institution of Civil Engineers; Vice-President, 

Engineering News Publishing Co. 
Author of ' Modern Tunnel Practice." 



Professor of Railroad Engineering and Geodesy in Cornell University. 
Author of "A Textbook on Geodesy and Least Squares." 


Professor of Architecture, University of Illinois: Fellow of the American Institute of 

Architects and of the Western Association of Architects. 
Author of " Elementary Graphic Statics and the Construction of Trussed Roofs." 



Civil Engineer. 

Author of "The Civil Engineer's Pocketbook." 


Professor of Civil Engineering and Applied Mechanics, McGill University, Montreal. 
Author of "A Treatise on Hydraulics." 


Author of "Planning and Construction of High Office Buildings," "Architectural Iron 
and Steel, and Its Application in the Construction of Buildings," " Compound Riv- 
eted Girders," 'Skeleton Structures,' etc. 



Professor of Civil Engineering, University of Illinois. 

Author of "A Treatise on Masonry Construction," " Engineers' Surveying Instruments, 
Their Construction, Adjustment, and Use," ''Roads and Pavements." 

Authorities Consulted Continued 


Assistant Professor of Structural Engineering, Sheffield Scientific School, Yale University. 
Author of " Plane Surveying: A Textbook and Pocket Manual." 


Joint Author of "A Treatise on Concrete, Plain and Reinforced. 


Author of " Modern Plumbing, Steam and Hot- Water Heating." 


Consulting Architect and Structural Engineer; Fellow of the American Institute of 

Author of "Architect's and Builder's Pocketbook," " Building Construction and Super- 
intendence, Parti, Masons' Work; Part II, Carpenters' Work; Part III, Trussed Roof s 
and Roof Trusses," " Strength of Beams, Floors, and Roofs." 


Professor of Civil Engineering, Columbia University; Consulting Engineer; American 

Society of Civil Engineers; Institution of Civil Engineers. 
Author of " Elasticity and Resistance of the Materials of Engineering;" Joint Author of 

"The Design and Construction of Metallic Bridges." 


Formerly Professor of Civil Engineering in Union University. 
Author of " Land Surveying and Direct Leveling," " Higher Surveying." 


President of the Brooklyn Engineers' Club; American Society of Civil Engineers; Ameri- 
can Society of Municipal Improvements; Principal Assistant Engineer, Department 
of Highways, Brooklyn. 

Author of " Street Pavements and Street Paving Material." 


Civil Engineer; President, The Pacific Northwestern Society of Engineers; American 

Society of Civil Engineers. 
Author of " Ordinary Foundations." 



Editor of The Railway and Engineering Review; American Society of Civil Engineers. 
Author of "Notes on Track Construction and Maintenance." 


Late Professor of Engineering at the Virginia Military Institute. 
Author of "A Treatise on Civil Ensrineerin.-r." 


HE marvelous developments of the present day in the field 
of Civil Engineering, as seen in the extension of railroad 
lines, the improvement of highways and waterways, the 
increasing application of steel and reinforced concrete 
to construction work, the development of water power 
and irrigation projects, etc., have created a distinct necessity 
for an authoritative work of general reference embodying the 
results and methods of the latest engineering achievement. 
The Cyclopedia of Civil Engineering is designed to fill this 
acknowledged need. 

C. The aim of the publishers has been to create a work which, 
while adequate to meet all demands of the technically trained 
expert, will appeal equally to the self-taught practical man, 
who, as a result of the unavoidable conditions of his environ- 
ment, may be denied the advantages of training at a resident 
technical school. The Cyclopedia covers not only the funda- 
mentals that underlie all civil engineering, but their application 
to all types of engineering problems; and, by placing the reader 
in direct contact with the experience of teachers fresh from 
practical work, furnishes him that adjustment to advanced 
modern needs and conditions which is a necessity even to the 
technical graduate. 

L The Cyclopedia of Civil Engineering is a compilation of 
representative Instruction Books of the American School of Cor- 
respondence, and is based upon the method which this school 
has developed and effectively used for many years in teaching 
the principles and practice of engineering in its different 
branches. The success attained by this institution as a factor 
in the machinery of modern technical education is in itself the 
best possible guarantee for the present work. 

L Therefore, while these volumes are a marked innovation in 
technical literature representing, as they do, the best ideas and 
methods of a large number of different authors, each an ac- 
knowledged authority in his work they are by no means an 
experiment, but are in fact based on what long experience has 
demonstrated to be the best method yet devised for the educa- 
tion of the busy workingman. They have been prepared only 
after the most careful study of modern needs as developed 
under conditions of actual practice at engineering headquarters 
and in the field. 

C. Grateful acknowledgment is due the corps of authors and 
collaborators engineers of wide practical experience, and 
teachers of well-recognized ability without whose co-opera- 
tion this work would have been impossible. 

Table of Contents 

BRIDGE ENGINEERING By Frank O. Dufour\ Page *11 

Bridge Analysis Early Bridges Trusses Girders Truss and Girder Bridges 

Deck and Through Bridges Truss Members Lateral Bracing Portals 
Sway Bracing Knee-Braces Kinds of Trusses Chord and Web Character- 
istics Weights of Bridges Loads (Dead, Live, Wind) Principles of Analysis 

Resolution of Forces Method of Moments Stresses in Web and Chord 
Members Warren Truss under Live and Dead Loads Position of Live Load 
for Maximum Moments Counters Maximum and Minimum Stresses Pratt, 
Howe, Baltimore, Bowstring, Parabolic, and Other Trusses under Dead and Live 
Loads Engine Loads Position of Wheel Loads for Maximum Shear and Mo- 
ments Pratt Truss under Engine Loads Impact Stresses Snow- Load 
Stresses Wind-Load Effects Top and Bottom Lateral Bracing Overturning 

Pratt Truss under Wind Loads Girder Spans Floor- Beam Reactions 
Plate-Girder Reactions Bridge Design Economic Considerations Types of 
Bridges for Various Spans Economic Proportions of Members Clearance 
Diagram Weights and Loadings Specifications Stress Sheet Design of a 
Plate-Girder Railway Span Masonry Plan Determination of Span Ties and 
Guard-Rails Web and Flanges Rivet Spacing Lateral Systems and Cross- 
Frames Stiff eners Web-Splice Bearings Design of a Through Pratt Rail- 
way Span Stringers Floor- Beams Tension Members Intermediate Posts 
Lacing Bars End-Posts Pins Transverse Bracing Shoes and Roller Nests 

HIGHWAY CONSTRUCTION . By A. T.Byrne and A.E.Phillips Page 267 

Country Roads Road Resistances to Traction Axle Friction Air Resistance 

Tractive Power and Gradients Effects of Springs on Vehicles Location of 
Roads Contour Lines Levels Cross- Levels Bridge Sites Mountain Roads 

Alignment Zigzags Construction Profile Width and Transverse Contour 

Drainage Ditches and Culverts Earthwork Roads on Rocky Slopes Earth 
and Sand Roads Grading Tools Rollers Sprinkling Carts Road Coverings 

Gravel Roads Macadam Roads City Streets Catch-Basins Pavement 
Foundations Stone- Block Pavements Properties of Stones Cobblestone 
Pavement Belgian Block Pavement Brick Pavement Paving Brick Con- 
crete Mixers Gravel Heaters Melting Furnaces Wood Pavements Asphalt 
Pavements Mixing Formulae Footpaths Curbstones Artificial Stone 
Pavement Selection Safety Life of Pavements Cost Relative Economies 


INDEX Page 405 

*For page numbers, see foot of pages. 

tFor professional standing of authors, see list of Authors and Collaborators at 
front of volume. 




1. Introduction. The following treatment of the subject of 
Bridge Analysis, while not exhaustive, is regarded as sufficiently 
elaborate to develop and instill the principal theoretical considera- 
tions, to illustrate the most convenient and practical methods of 
analyzing the common forms of trusses and girders, and also to lay 
a sufficient foundation for the analysis of such other trusses as are 
not specifically mentioned or treated herein. 

The necessary steps and operations required for a proper analysis 
of the several types of bridges are fully demonstrated by sketches 
and computations, the numerical values being mechanically obtained 
by the use of a slide rule, which is a handy instrument for quickly 
performing the operations of multiplication and division, and for 
squaring and extracting the roots of numbers. The values given 
may differ from the exact value by one unit in the second decimal 
place (seldom more) and are sufficiently accurate for the purpose of 
design. All bridge computers should be proficient in the use of the 
slide rule. 

The problems given in the back of this Instruction Paper, 
exemplifying the practical application of the subject-matter treated 
in the various articles, should be solved by the student as each article 
is mastered. 


2. Early Bridges. Early bridges were not bridges according 
to the present conception of the term. They were simple pile 
trestle bents placed at frequent intervals and connected by wooden 
beams on which the floor was placed. The Pons sublicius, built over 
the Tiber, at Rome, about 650 years before Christ was born, was of 
this trestle type. Also the famous bridge b'uilt by Caesar across the 
Rhine in 55 B. C. was of the same kind of construction. As civiliza- 
tion progressed, the arch type was developed; and in 1390 the great 

Copyright, 1908, by American School of Corespondence. 



bridge at Trezzo over the River Adda was built of one span of 251 
feet, which has never been eclipsed or equaled. 

3. Truss Bridge Development. The first truss bridge is sup- 
posed to have been originated by Palladio, an Italian, who used the 
king-post truss (Fig. 1) about 1570. Its importance was not recog- 
nized, and it became entirely for- 
gotten until it was rediscovered in 
1798 by Theodore Burr, an Ameri- 
can, who used it in his construction. 
About the same time, Burr invented 
the truss that bears his name, 
which was in reality a series of 
king-post trusses (see Fig. 8). This 
was found to be unstable under moving loads, and was therefore 
stiffened by the use of an arch (Fig. 2), or was built somewhat as an 
arch, there being considerable rise at the center of the span (Fig. 3). 
By 1830 the principle of the double cross-bracing in the panel was 
understood; and in quick succession came the patents of Long, 
Howe, Pratt, and Whipple on forms of trusses which bear their 
respective names. 

It remained for Squire Whipple in 1847 to place the science of 
bridge building on a rational and exact mathematical basis such 

Fig. l. King-Post Truss. 

Fig. 2. King-Post Truss Bridge Stiffened by Arch. 

as is now used. Previous to this time, and indeed several years 
afterwards for Whipple's work did not become generally known 
until a much later date bridges were built, not from previously 
computed strains, but by "judgment." All parts of a bridge were 
made of the same size, and if one started to fail it was replaced by 
a larger one; or small models were made and loaded proportionally, 
broken members being replaced by larger ones. There is no doubt 



that many of the bridges built at this period were very weak as well 
as very strong. The failures are not remembered; but the sound 
judgment of many of our earlier bridge engineers is recorded in the 
wooden structures they left behind them, some of which have stood 
the demands of traffic for over a century. After 1850, bridges were 
built from computed stresses; wood was discarded; and the develop- 
ment became rapid, until about 1870, when the introduction of sub- 
diagonal systems brought the truss system to practically what it is 


4. Trusses. A truss is a series of members taking stress in 
the direction of their length, placed together so as to form a triangle 

Fig. 3. Burr Trusts Bridge, Arched. 

or system of triangles, which, when placed upon supports a certain 
distance apart, will, in addition to their own weight, sustain certain 
loads applied at the points where the members intersect. These 
points are called panel points, 

5. Bridge Trusses. A bridge truss is one in which the members 
that carry the superimposed loads are in the same plane. Usually 
this plane is vertical. 

6. Truss Bridges. A truss bridge is a structure consisting of 
two or more usually two bridge trusses connected by a system 
of beams called the floor system, which transfer to panel points the 
load for which the trusses are designed. 

7. Girders. These are beams consisting of a wide, thin plate, 
called a web plate, with shapes, usually angles and narrow, thin 
plates called flanges, at the top and bottom edges. All are firmly 
riveted together. (See Part IV, "Steel Construction.") 

8. Girder Bridges. These consist of usually two, sometimes 



three, girders connected as in the case of truss bridges by a system 
of beams. 

9. Deck Bridges. In cases where the floor system connects 
the trusses at their tops, the bridge is called a deck bridge, since the 
traffic moves on a deck, as it were (see Fig. 4). 

Fig. 5. Through Bridge. 

10. Through Bridges. In cases where the floor system con- 
nects the bottoms of the trusses, the bridge is called a through bridge, 
as the traffic moves through the space between the trusses (see Fig. 5). 

11. Members of a Truss. Each truss consists of a top and 
bottom chord, end-posts, and web members. The web members are 
further divided into hip verticals, intermediate posts, and diagonals. 
Fig. 6 shows these various classes, A- A being top chord, B-B 

Fig. 6. Showing the Members of a Truss. 

bottom chord, A-B end-posts, vertical members C-b intermediate 
posts, A-a hip verticals, and A-b and C-b diagonals. 

12. Pony=Truss Bridges. When the height of the trusses of 
a through bridge is less than the height of the loads that go over 
them, they are called pony trusses, and the bridge a pony-truss bridge. 



13. Lateral Bracing. In all deck bridges, and in all through 
bridges except pony-truss bridges, the chords which are not con- 
nected by the floor system are connected by a horizontal truss system 
called the lateral bracing. In all bridges the chords which are con- 
nected by the floor system are connected by a horizontal truss system, 
also called the lateral bracing. One of these systems is called the 


Fig. 7. Through Bridge. Showing Top and Bottom Systems of Lateral Bracing, also 
Portal Bracing and Floor System. 

top lateral system, as it connects the top chords; and the other is called 
the bottom lateral system, as it connects the bottom chords (see Fig. 7). 

14. Portals. In through bridges, the end-posts of the pair 
cf trusses are connected by a system of braces in order to preserve 
the rectangular cross-section of the bridge. This is called the portal 
bracing (see Fig. 7). 

15. Sway Bracing and Knee=Braces. These serve the same 
purpose as the portal braces, and are either small struts or systems 



of cross-bracing placed at the intermediate posts. The former are 
called knee-braces, and the latter sway bracing. 

16. Floor Systems. In both highway and railway bridges, 
there are beams running from the intermediate posts or hip ver- 
ticals across to the like members opposite. These are called floor- 
beams. In highway bridges, there are smaller beams running parallel 
to the trusses and resting at their ends upon the floor-beams. These 
are called floor-joists, and the plank or other floor rests directly upon 
them. In railway bridges, two beams or girders per track run parallel 
to the trusses and are connected at their ends to the floor-beams. 
These are called track stringers (or simply stringers'). The ties rest 
directly upon them. The various members of the floor system of 
a railway bridge are shown in Fig. 7. The diagonals connecting the 
top chords, and those connecting the bottom chords, are the top and 
bottom laterals respectively. 


17. Names. Trusses may be classified according to their 
names, the character of their chords, and the system of w r ebbing. 
Table I gives the classification of the more important of these accord- 
ing to name. 

Chronological List of Trusses 









Fig. 1 



Theodore Bun- 


Fig. 1 



Theodore Burr 


Fig. 8 




Fig. 9 



William Howe 


Fig. 10 



Thos & Caleb Pratt 


Fig. 11 



Squire Whipple 
Squire Whipple 
Penn. R. R. 


Fig. 12 
Fig. 13 
Fig. 14 

Of the types of trusses listed in Table I, the Warren, Howe, 
Pratt, Bowstring, and Baltimore are now built; and of these construc- 
tions probably 90 per cent are Pratt trusses. The Baltimore truss 
is used for long spans only. 

18. Chord Characteristics. In most types of bridges the 



Fig. 8. Burr Truss. 

Fig. 9. Warren Truss. 

Fig. 10. Howe Truss. 

Fig. 11. Pratt Truss. 

Fig. 12. Whipple Truss. 



chords are parallel. When such is the case, the stresses increase 
from the end toward the center, and there is a considerable difference 
between any two adjacent panels of the same chord. This neces- 
sitates different areas for each section. When the chords are not 
parallel, as in the bowstring truss, the stresses in the chords are so 
nearly equal that the same area is used throughout or nearly through- 
out the entire chord. Also, the stresses in the diagonals are nearly 
equal. These conditions would seem to indicate that this was a very 
economical form of truss. Theoretically it is; but practical con- 
siderations such as the beveled joints and the posts which must be 
constructed to withstand reversals of stress customarily limit this 
type to the longer spans. 

19. Web Characteristics. The web systems of the Burr, 
Warren, Howe, Pratt, and Bowstring trusses are called single sys- 
tems; that of the Whipple truss is a multiple system; while those of 
the Baltimore trusses are examples of webbing with sub-systems. 
As the maximum economical panel length has been found to be about 
25 feet, which makes the economical height of the truss about 30 
feet, and as the length of the span should not be more than ten times 
the depth, the span for trusses with simple systems of webbing is 
limited to about 300 feet. In order to increase the limiting span, 
multiple systems like that of the Whipple or similar ones were intro- 
duced. Calculations of stresses in members of the Whipple truss are 
somewhat unreliable on account of the fact that we are unable to 
tell just how the effects of the loads are distributed. For this reason, 
that type has gone out of use, and the sub-systems are used instead. 
These allow spans of twice the above limit; and, indeed, trusses with 
this type of webbing have been built up to and over 600 feet. This 
style of webbing can be applied to the bowstring truss, almost all 
long-span bridges being of this type with sub-systems of webbing. 


20. Formulae. In order to obtain the stresses due to the 
weight of the structure, the latter quantity must be known. As 
this weight can be determined only after the structure has been 
designed, it is evident that an assumption as to the weight must be 
made. The best method is to use the actual weight of a similar 
structure of like span which has been built. As the necessary data 


for this is not always available, it is customary to use formulae to 
derive an approximate weight of sufficient accuracy for purposes of 
computation. Table II gives some of the most reliable formulee. 

Formulae for Weights of Bridges 




Heavy Interurban 


w=6QQ + l.&l + 27b + -Lbl(l + -rjm 1 ) 

E. S. Shaw 

First-Class High- 

/ \ 

way Riveted 

M) = 300 + l + 22b-\ bl(l H 1 ) 

E. S. Shaw 

First-Class High- 

15 \ 1,000 / 

way Pin 

w = 34 + 226 + Q.lQbl + 0.71 


Light Country 

w = 250 + 2.51 


Railroad Truss 

E 50 

w = (650 + 70 

F. E. Turneaure 

Railroad Truss 

E 40 

w = | (650 + 70 

F.E. Turneaure 

Railroad Truss 

E 30 

w >> | (650 + 70 

F E. Turneaure 

Railroad Deck 

Girder E 50 

w = 124.0 + 12.01 


Railroad Deck 

Girder E 40 

w = 123.5 4- 10.01 


Railroad Deck 

Girder E 30 

w = 111.0 + 8.81 lAuthor 

In the above formulae, w = Weight of steel per linear foot of span; 
I = Length of span in feet; b = Breadth of roadway, including sidewalks. 

In using the formulae of Table II, remember that a span has two trusses. 
The weights for highway bridges do not include the weight of the wooden 
floor, which may be assumed as 10 pounds per square foot of floor surface 
All highway bridges have steel joists. The weights of railroad spans do not 
include the weight of the ties and rails, which may be assumed at 400 
pounds per track per linear foot of span. If solid steel floors are to be used, 
700 pounds per linear foot of span are to be added to the weights computed 
from the table. 

All the weights given for railroad spans are for single track. Double- 
track truss-spans are about 95 per cent'heavier; and double-track girder-spans 
are 100 per cent heavier. Through girder spans are about 25 per cent heavier 
than deck girder spans; and through spans are about 15 to 40 per cent heavier 
than deck spans. 

The spans on which Table II is based are of medium steel. Bridges 
built of soft steel or wrought iron will weigh 10 to 15 per cent more. 

* The author is indebted to the distinguished engineers whose names appear in 
Table II, for permission in this connection to make use of the formulas given opposite 
their names. 




In order to give an idea of the relative weights of steel in different 
classes of bridges, let it be required to compute the dead weight of a 
100-foot span of each class. For heaviest highway bridges to carry 
heavy interurban cars: 
u> = 600-f!80+27 X16 + 16 *100 t + _HH = i 358 lbs . per linear ft . 

Fig. 13. Bowstring Truss. 

For heavy riveted highway bridges to carry heavy farm engines : 

u = 300 + 100 + 22 X 16 + 16 100 1 + - = 870 Ibs. per linear foot. 

Fig. 14. Two Forms of Baltimore Trusses. 

For heavy pin-connected highway bridges to carry heavy farm or 
traction engines: 

w = 34 +22 X 16 + 0.16 X 16 X 100 + 0.7 X 100 = 710 Ibs. per linear ft. 
For light country highway bridges to carry 100 pounds per square 
foot of floor surface: 

w = 250 + 2.5 X 100 = 500 Ibs. per linear foot. 
If the total weight is required, the weight of the wooden floor 
must be added. Take, for example, the last bridge: 



Weight of steel = 500 X 100 = 50 000 pounds. 

" floor = 100 X 16 X 10 = 16000 pounds. 

Total dead load = 66 000 pounds. 

The weight per linear foot for a railroad truss bridge of 100-foot 

span is: 

w = 650 + 7 X 100 = 1 350 Ibs. per linear foot. 

This is about the same as that for a heavy interurban bridge 
The reason for this is that in addition to the heavy rolling stock 
of the electric road, the heavy highway traffic must be provided for. 
A deck girder of 100-foot span weighs: 

w = 124 + 12 X 100 = 1 324 Ibs. per linear foot. 

21. Actual Weights of Railroad Spans. In case actual weights 
can be obtained, a more exact analysis can be made. The weights 
of bridges indicated in the accompanying tables and diagrams, are 
based on actual constructions recently erected. These bridges rep- 
resent the very best modern practice of engineers and manufacturers 

The weights of through truss-spans made of medium steel 
and designed for E 50 loading, are given in Fig. 15. The weights 
include the weight of the ordinary open steel floor, and they also 
include the weight of the ties and rails, which is taken at 400 pounds 
per linear foot per track. 

The weight of steel in medium steel deck plate-girder spans 
designed for E 50 loading, is given in Table III. 


Weights of Deck Plate-Girders, Medium Steel 

Loading E 50 

(in feet) 

(in pounds) 

(in feet) 

(in pounds) 










11 800 










105 500 






27 400 


123 600 


32 400 






161 700 




174 900 


51 500 



The spans are the distance center to center of bearings; and the weights 
do not include the weight of the ties and rails, which is to be taken at 400 
pounds per linear foot per track. Intermediate spans may be interpolated. 



22. Actual Weights of Highway Spans. The actual weights 
of highway spans for heavy interurban trolley-cars and traffic, should 
preferably be computed from the formulae of Shaw or Waddell (Table 
II). ,The weights of country bridges, including floor, may be taken 
from the diagram of Fig. 16. 


23. Classes of Loads. Those weights just given constitute 
what is called the dead load of the bridge. The traffic which passes 




























A - / 




Fig. 15. Weights of Through Truss-Spans. Fig. 16. 
Medium Steel, E 50 Loading. 

Weights of Country Bridges, In 
eluding Floor. 

over the bridge is called the live or moving load. In addition to the 
two classes mentioned, is the effect of the wind, which is designated 
as the wind load. These loads vary with the class of bridge, be it 
highway or railway, and with the purpose for which it is intended. 

24. Live Loads for Highway Bridges. Highway bridges are 
usually divided into several classes according to the traffic, which 
may be that of heavy interurban cars, light trolley-cars, farm engines, 
road rollers, teams, human beings, or some combination of these 
loadings. The standard specifications of J. A. L. Waddell or of 
Theodore Cooper are obtainable for a very small sum. Their pur- 


chase is advised, and the reader is referred to them for further infor- 

The trusses of country highway bridges are usually designed for 
a live load of 100 pounds per square foot of roadway. This may be 
considered good practice; and it is the law in some States. The 
floor system of these same bridges should be of sufficient strength to 
sustain 100 pounds per square foot of roadway, or a 12-ton farm 
engine having 4 tons on the two rear wheels, which are 12 inches wide 
and 6 feet apart, and 2 tons on each of the front wheels, which are 6 
inches wide and 5 feet apart. The axles of this engine are spaced 
8 feet center to center. 

25. Live Loads for Railway Bridges. The loads for any par- 
ticular railroad bridge are not always the same, on account of the 
great variation in the weights and w r heel spacings of engines and 
cars. It is customary to design the bridge for the heaviest in use 
at the time of construction, or for the heaviest that could reasonably 
be expected to be built thereafter. 

As the computations with engines were formerly somewhat 
laborious on account of the different weights and spacing of wheels, 
it has been proposed by some engineers to use a uniform load, called 
the equivalent load, which would give stresses the same, or very nearly 
the same, as those obtained by the use of engine loads. However, 
as these loads are different for each weight of engine, and also different 
for the chord members, the web members, and the floor-beam reaction 
of each different length of span, and as the labor of the computations, 
using engine- wheel loads, has been greatly reduced by means of 
diagrams, it does not seem as if this method would ever come into 
very general favor except for long-span bridges, where the live load is 
much smaller than the dead load. 

The equivalent loads for Cooper's E 40 (see Fig. 85) are given 
in Table IV. 

Most railways specify that their bridges shall be computed by 
using two engines and tenders followed by a train. The spacing 
of the wheels, and the load which comes on each wheel of the engines 
and tenders, are fixed by the railway company. The train is repre- 
sented by a uniform load. Formerly there was a great diversity of 
practice among the different roads in regard to the engine and. train 
loads specified; but practice has of late years become quite uniform, 



Equivalent Uniform Loads 

Loading E 40 

(in feet) 


(in feet) 









9 000 




6 330 





11 640 




7 140 



9 340 

11 330 



6 110 

" 7060 

5 140 


9 340 

11 080 





5 130 



10 860 



5 960 i 6 820 



9 030 

10 670 




6 720 

5 110 



10 500 








10 350 








10 240 

6 780 







10 100 








10 000 

6 500 







9 780 

6 390 








6 290 
















6 120 



6 300 












































8 430 




6 180 








6 150 



7 140 

8 190 




6 130 








6 110 













5 570 


5 170 








5 150 








5 140 | 6 020 







5 100 










4 700 

















44 6 470 7 340 






with an apparent tendency to standardize in accordance with the 
classes of loading specified by Cooper. Cooper's Class E 50, which 
represents the heaviest engines now in common use, was invented 
by Theodore Cooper, a consulting engineer of New York City. It is 
given in Fig. 17. 

Lighter loadings for light traffic on the same general plan are 
advocated by Mr. Cooper, and are given at length in his "General 
Specifications for Iron and Steel Railway Bridges and Viaducts" 
(1906 edition). 

26. Wind Loads. Some designers require that the stresses due 





0~OOOQ (^ 

ooffos \^^ ~r 

to wind shall be computed by using 30 pounds 
per square foot of actual truss surface. This 
requires that you know the size of the mem- 
bers of the bridge before it is designed which 
is evidently an impossibility; or that an as- 
sumption as to their size be made which 
allows a chance for a mistake in judgment, 
especially in an inexperienced person. A more 
logical method, and one used to a great extent, 
is to assume a force of so many pounds per 
linear foot to act on the top and bottom chords 
and on the traffic as it moves across the bridge. 

In highway through bridges, it is the usual 
practice to take the wind load as 150 pounds 
per linear foot of top and bottom chords, and 
150 pounds per linear foot of the amount of 
live load which is on the bridge. 

For railroad bridges, it is customary to use 
considerably higher values than those used in 
highway practice not that the wind blows 
harder on railroad than on highway bridges, 
but so that the bracing designed by the use of 
these values may be sufficiently strong to stiffen- 
the bridge not only against the wind, but also 
against the vibrations caused by the rapidly 
moving traffic. Good practice for through 
bridges is to use 150 pounds per linear foot 
of the top chord, 150 pounds per linear foot 
of the bottom chord, and 450 pounds per linear 
foot of live load on the bridge. This latter 
force is supposed to act at a line 8.5 feet above 
the base of the rail. 

For deck bridges, for both highway and 
railway use, the unit-loads on the moving or live 
load remain the same, but the unit-loads on 
the top and bottom chords are reversed. 

In computations involving the live load, it 
is always assumed that the live load moves over 
the bridge from right to left. 




27. Principles of Analysis. The stresses in bridge trusses may 
be determined by both algebraic and graphic methods. In some 

P P P p p 

a R 

Fig. 18. Truss under Loads, Illustrating Principles of Analysis. 

cases, one is more expeditious than the other. Algebraic methods 
alone will be considered in this text. 

The analysis of stresses is based upon the fact that the interior 
stresses in a member or group of members hold in equilibrium the 
exterior forces. That this is a fact, can easily be understood. Con- 
sider a man pulling on a rope which is fastened at one end to an im- 
movable object. There will be a 
stress in the rope equal to, and 
opposite in direction to, the pull 
exerted by the man. In order to 
prove this, cut the rope and ap- 
ply a force equal and opposite to 
the pull exerted by the man, where 
the cut is made ; and the rope 
and man will be in equilibrium. 
Also, suppose that a truss under 
loads, as indicated by the arrows, 
Fig. 18, were cut along the section 
a-a, and that forces F y F 3 , F 5 
equal to the stresses S 2 , S 3 , and S 5 were placed at the ends of the 
members as indicated in Fig. 19, then that portion of the truss to 
the left of the section would be in equilibrium. The interior stresses, 
represented by F 2 , F 3 , and F 5 , would hold in equilibrium the exterior 
forces p and R. 


Forces Substituted for Stresses in 
Truss of Fig. 18. 




From inspection of Fig. 19, it will appear evident that, as the 
position of the truss to the left of the section is in equilibrium, the 
following statements are true : 

1. The algebraic sum of the moments of the exterior forces and the 
stresses in the members cut by the section, is equal to zero. This is true of the 
moments taken about any or all points; for, if it were not, the portion of the 
truss would begin to rotate about some point, and would continue until 
equilibrium was established. 

2. In a vertical plane, the algebraic sum of the components of the 
exterior forces and the stresses in the members cut by the section is equal 
to zero; for, if such were not the case, the portion of the truss shown would 
move up or down with a constant acceleration. 

3. The algebraic sum of the components of the exterior forces and the 
stresses in the members cut by the section in a horizontal plane, is equal to 
zero; for, if such were not the case, the portion of the truss would move either 
to the right or to the left, with a constant acceleration. 

4. From 2 and 3, above, it is evident that the algebraic sum of the 
components of the exterior forces and the stresses in the members cut by 
the section is equal to zero in any and all planes. 

The section is not necessarily required to be a vertical line as in 

Fig. 20. 

Oblique Section Cutting 

Fig. 20a. 

Circular Section 
Cutting Truss. 

Fig. 21. 

Illustrating Resolution 
of Forces. 

Fig. 19. It may be oblique, as in Fig. 20; or it may be a circular 
section, as shown in Fig. 20a. When the latter is the case, it is said 
that the sum of the components of the forces around the point U l 
is in equilibrium in any plane that may be taken. 

It is also evident that the forces in the members cut by the 
section, and the exterior forces to the right, are in equilibrium. 
This condition is very seldom utilized in the determination of stresses, 
as that portion of the truss to the left of the section is almost always 

28. Resolution of Forces. This method is one of the simplest 




and at the same time least laborious. The forces are generally 
resolved into their horizontal and vertical components, or parallel 
and perpendicular to some member. In cases where two unknown 
stresses occur, two equations can usually be formed, and these solved. 
It should be assumed that the unknown stress acts away from 
the section which cuts it. It will then solve out, with the proper 
sign indicating the character of the stress that is, whether it is 
tensile or compressive. Tensile stresses are indicated by placing 

6. P P 

Fig. 24. Pig. 25. 

Diagrams Illustrating Application of Method of Resolution of Forces in Analysis of 

the plus ( + ) sign before them, while a minus ( ) sign indicates 

A few equations showing the application of the method of the 
resolution of forces can be written after inspection of Figs. 21 to 25 
inclusive. In all cases, S x is the unknown stress, and is assumed 
to be acting away from the section. The other stresses S 1} S 2 , etc., 
are known, and their direction given them accordingly, it being 
toward the section if the member is in compression, and away from 
the section if the member is in tension. Forces or components 
acting upward or to the right are considered plus ; those acting down- 


ward or to the left are considered minus. For a fuller explanation, 
see the instruction paper on Statics, Articles 17 to 23 inclusive. 

In Fig 21, the sum of the vertical components is taken, and the 

equation is : 

+ R - p - p - S* cos (j) = 0; 

<S X = ( + R p p) sec <f>. 

In Fig. 22, the section is oblique, and the sum of the vertical 
components is taken : 

+ R - p - p + S x = 0; 

S x = - ( + R - p - p). 

In both of the above cases, it will be noted that the chord 
stresses do not enter into the equation, as their vertical components 
are zero. 

In Fig. 23, the sum of the horizontal forces is used in deter- 
mining the stress <S X . Note that the exterior forces R and p do not 
enter the equation, as they are not to the left of the section, and also 
their horizontal components are zero. 

+ S l sin $ + S 2 sin + S 3 sin <j> + S x = 0; 

S K = - (S l + S 2 + S 3 ) sin <f>. 

In Fig. 24, the sum of the vertical forces is again used. Here 
the section cuts the member with the known tensile stress S r 

+ R p p - p p - p + S l cos (j) - S x cos $ = 0; 

S x = + (R - 5p) sec $ + Si. 

In Fig. 25, use is made of the fact that the sum of the components 
of the forces about a point is zero when they are resolved in any plane. 
Here they will be resolved perpendicular to the diagonal. 

-S,sin-0 + S K = 0. 

S x = + S 1 sin <. 

These are some of the most common cases which occur in the 
determination of stresses in simple trusses. In all cases, follow this 
method of procedure: 

1. Pass a section cutting as few members as possible, one of which 
must be the one whose stress is desired. 

2. The stress in all the members cut, with but one exception, must be 

3. Write your equation, always placing it equal to zero. 

4. Solve for your stress. 




29. Method of Moments. The stresses in all members of a 
truss can be determined by this method. By section 1 of Art. 28, 
the point about which the moments are considered can be taken 
anywhere. Fig. 26 represents the point as taken somewhere outside 
of the truss at a distance a above the point V \. The equation will 

then be: 

-S, X a - S 2 X b - S 3 (a + A) + Rp P, X + P 2 X p = 0. 

This involves three unknown quantities, and therefore two other 

points should be taken, and two 
more equations written. By the 
use of the three equations, the 
stresses can be determined. 

It is customary to assume the 
center of moments at such a place 
that the moments of all the un- 
known stresses, with one excep- 
tion, are zero. This condition 
requires that their lines of action 
pass through the center of mo- 
ments. Let it be required to 
- 5 3 determine the stress *S 3 . If the 
center of moments is taken at 
U v then, as the lines of action 
of <Sj and S 2 pass through this 

Fig. 26. Diagram illustrating Application point, their moments will be zero, 

of Method of Moments in Analysis 

and the following is true: 
+ R X 2p - P, X p P 2 X - S 3 X h = 0. 


S, =-- 

X 2p - P, X p). 

Likewise, if the top chord is curved, the center of moments can be 
taken in such a position that only the unknown stress will enter into 
the equation. If it is desired to determine the stress S 2 , Fig. 27, 
the equation would be : 

-S 2 X I - R X a + P, (a . + p) + P 2 (a + 2p) =0, 
the center of moment being at 0, the intersection of the lines of stress 
of S l and S y Solving the equation just stated, 

S 2 = -L j -Ra + P, (a + p) + P 2 (a + 2p) 1 




30. Stresses in Web Members. By reference to Articles 28 and 
29, it is seen that several methods are presented for the solution of 
stresses in web members. Each should be adapted to the case in 
hand. The simplest method, and the one which is commonly used 
in all trusses with parallel chords, is by the resolution of the vertical 
forces. Fig. 21 is to be referred to. The equation given on page 

19 is: 

+ R - p p - Sx cos $ = 0. 

Fig. 27. Diagram Illustrating Application of Method of Moments In Analysis of Trusses. 
Top chord curved. . 

But R p p is equal to V, the vertical shear at the section, and 
so the equation may now be written : 

V - S x cos<f> = (1) 

whence the following important rule is deduced: 

'The algebraic sum of the vertical shear at the section and the ver- 
tical components of the stress in all of the members cut by the section, is 
equal to zero. 

In trusses with horizontal chords and a simple system of webbing, 
the equation may be put in the form: 

S* = + V sec <f>; 

and the statement that the stress in any web member is equal to the 
shear times the secant of the angle that it makes with the vertical is 




true. The practice of using this latter statement is not to be en- 
couraged, as it leads to confusion in the signs of the stresses. Equa- 
tion (1 ) should be written in all cases, and the stress will then solve with 
its correct characteristic sign, indicating that the stress is either 
tensile or compressive. 

As an example, let it be required to determine the stresses in the 
web members S 2 and S 3 of the Pratt truss shown in Fig. 28, the loads 
being in thousands of pounds. First, a section should be passed, 
cutting that member and as few others as possible. Next, the shear 
at that section should be computed. Then the vertical components 
of all the stresses cut by the section, and the vertical shear, should be 

Fig. 28. Calculation of Stresses in Web Members of Pratt Truss. 

equated to zero. Finally, solve the equation. Remember that the 
unknown stress is to be assumed as acting away from the section, 
and that forces or resultants acting downward are considered negative, 
while those acting upward are considered positive. 
To determine S 2 : 

The vertical shear at the section a a is : 

+ 37.5 -2 X 10 - 5 = +12.5. 
As the chord stresses do not exert a vertical component, the equation is: 

+ 12.5 + S 2 = 

S 2 = 12.5, which is a compressive stress of 12,500 pounds. 
Note that in this case the angle which the member makes with the 
vertical is zero, and the cosine and secant are unity. 
To determine S 3 : 

The vertical shear at the section b b is 

+ 37.5 -2X10-2X5= +7.5. 
The equation is : 

+ 7.5 - S 3 cos = 

S 3 = +7.5 -sec <. 



Sec </> is equal to V 3Q 2 + 25 2 H- 30, which is equal to 1.302; and 

S 3 = +7.5 X 1.302 

= +9.765, which is a tensile stress of 9,765 pounds. 

31. Stresses in Chord Members. The stresses in chord may be 
obtained by either the method of moments or the method of resolu- 
tion of forces, this latter being usually the resolution of horizontal 

In accordance with the text of Article 29, the following rule may 
be stated with regard to the solution of stresses in chord members 
by the method of moments : 

Pass a plane section cutting the member whose stress is to be computed, 
and as few others as possible; then take the center of moments at such a point 

/ I \ \ x ' , , \ 

i- V -A 

Pig. 29. Calculation of Stresses in Chord Members by "Tangent" or "Chord-Increment" 


that the lines of action of as many forces as possible, the unknown one excepted, 
pass through that point; write an equation of the moments about this point of 
the known loads and forces to the left of the section, assuming the unknown force 
to act away from the section, and taking the known forces to act as given, the 
tensile stresses to act away from the section, and the compressive stresses to act 
towards the section; place the equation equal to zero, and solve. 

The stress will solve out with its correct characteristic sign. 

In the majority of cases a section can be made to cut three mem- 
bers only, one of the three being the one whose unknown stress is 
desired. In such cases, take the center of moments at the inter- 
section of the other two, and proceed as before. As examples of this 
latter case, note the centers of moments at U 2 , Fig. 26, and 0, Fig. 
27, and also the equations resulting therefrom. 

When the method of resolution of forces is used, it is usually 
designated as the tangent method or the chord increment method. The 
simplest application of this method is to trusses with horizontal 
chords and vertical posts in the web members. Then the stress in 
any chord member is equal to the product of the sum of the shears 


in the panels up to that section, and the tangent of the angle which 
the diagonals make with the vertical. 

This can readily be proved by reference to Fig. 29. Let it be re- 
quired to determine the stress in the chord member S x . Pass the 
section a a. The stresses S v S 2 , <S 3 , and S 4 are now computed, 
and are S t = - V l sec <f>; S 2 = +V 2 sec <f>; S 3 = +V 3 sec <f>; and S 4 = 
+ F 4 sec <. Now noting the directions of the known stresses and 

Lo L, L E L 3 L* L 5 L 6 

Fig. 30. Illustrating Method of Notation of Stresses and Members in a Through Bridge. 

LI Lg l_3 I_4. I_5 

Fig. 31. Illustrating Method of Notation of Stresses and Members in a Deck Bridge. 

assuming S x to act away from the section, the equation of the hori- 
zontal component is: 

+ <S, sin $ + S 2 sin $ + S sin <}> + S t sin < + S x =0. 
Now, substituting the values of S t , <S 2 , etc., and remembering that 
sec = - -JT, the equation becomes: 

. y Bin sin ^ Bin . 

. * cos ^ 2 cos ^> 3 cos 

from which, . 


4 COS 

r 3 + T 4 ) tan <; 

S x = -2"F tan <j>. 

From inspection of Fig. 29, it will be noticed that the stress in 
any section of the chord is equal to that in the section to the left of 



it, plus the increment (horizontal component) of the diagonal; hence 
the name chord increment method. 

32. Notation. The practice hitherto used in designating 
stresses by S v S 2 , etc., will now be discontinued, as it is inconvenient 
in the extreme; moreover, it is not the method used in practical work. 
The notation to be used is that given in Figs. 30 and 31, the former 
being for a through and the latter for a deck truss. 

The practical advantages of this system are very great. When 
U l U 2 is noted, it is at once known to be the top chord of 
the second panel; 7 2 L 2 is known to be the second vertical ; while U 2 
L 3 is at once recognized as the diagonal in the third panel. A stress 
in a member, as well as the member itself, is designated by the 
subscript letters at its ends. Thus U 1 L 2 may mean the member 

Fig. 32. Calculation of Stresses in a Six-Panel Warren Truss Through Bridge. 

itself or the stress in the member. The text will clear this up. In 
analysis, the stress would be implied, while in design the member 
itself would be intended. 

33. Warren Truss under Dead Loads. The Warren truss has 
its web members so built of angles and plates or of channels, that 
they can take either tension or compression. The top chord is of 
structural shapes, while the lower chord may be of built-up shapes 
or simply of bars. 

Let it be required to determine all of the stresses in the six- 
panel truss of a through W T arren highway 120-foot span for country 
traffic. The height is to be 20 feet. The outline is given in Fig. 32. 
According to Fig. 16, the total weight of the span, including wooden 
floor, is 76 000 pounds. Each truss carries one-half of this, or 76 000 


-7- 2 = 38 000 pounds. As there are six panels, each panel load is 
38 000 -7-6 = 6 333 pounds. This means that we must compute 
the stresses in the above truss by considering that a load of 6 333 
pounds is at points L v L 2 , L 3 , L 4 , and L 5 . Of course there is some 
weight at L and L 6 ; but this does not stress the bridge, as it is 
directed over the abutments or supports. The reactions at L and 
L 6 are each equal to (5 X 6 333) -=- 2 = 15 833 pounds. 
The shears are next computed, and are: 

V l = +15833 - = +15833 

F 2 = +15 833 - 6 333 = +9 500 

F 3 = + 15 833 - 2 X 6 333 = + 3 167 

It is unnecessary to go past the center of the bridge, as it is symmetri- 
cal. The V l represents the shear on any section between L and L x ; 
V 2 represents the shear on any section between L l and Z/ 2 ; and F 3 
represents the shear on any section between L 2 and L 3 . The secant 
of the angle <f> is : 

H- 20 = 1.12. 

The stresses in the web members are computed as follows: 
For L u 7 r Pass section a a. Assume stress acting away from 
the section, as shown. Then, 

F, + L C7, cos = 0; 

L U l = Fj sec 0; 

L U, = - 15 833 X 1.12 = - 17 700 pounds, 

which shows that L U l has a compressive stress of 17 700 pounds. 
For U l L r Pass section 6 6. Assume stress acting away from 
the section, as shown. Then, 

F, U l L l cos = 0; 

C/.L, = +T, sec 0; 

UiLt = +15833 X 1.12 =+ 17 700 pounds, 

which shows that UJL^ has a tensile stress of 17 700 pounds. 
For LJJ 2 . Pass section c c. Then, 

F 2 + LI C/ 2 cos = 0; 

Z^t/2 = F 2 sec 0; 

L,C7 2 = - 9500 X 1.12 = - 10 640 pounds. 

For U^L V Pass section d-d. Then, 

+ 9500 - U 2 L 2 cos = 0; 

UjL, = +9500 X 1.12 = +10640. 


For L 2 U y Pass section e e. Then, 

+ 3 167 + L 2 U 3 cos<f> = 0; 

L 2 U 3 = -3 167 X 1.12 = -3 540. 

For U 3 L 3 . Pass section / -/ . Then, 

+ 3 167 - U a L 3 cos < = 0; 

U 3 L 3 = +3 167 X 1.12 = +3 540. 

The computation of the stresses in the chords is made by the 
method of moments, and is as follows: 

For LJL^ Section b b cuts U l L l and UJJ 2 , besides the mem- 
ber whose stress is desired, and therefore the center of moments will 
be taken at their intersection f7 r The equation is : 

+ 15 833 X 10 - L Q L, X 20 = 0, 

L L, = ( + 15833 X 10) * 20; 

= +7 917 = a tension of 7 917 pounds. 

For LiL r Either section c c or d d may be used, and each 
shows the center of moments to be at U 2 . The equation is : 
+ 15 833 X 30 - 6 333 X 10 - L,L 2 X 20 = 0; 
L,L 2 = ( + 15833 X 30 - 6333 X 10) -H 20; 

= +20 583 = a tensile stress of 20 583 pounds. 

For L^Ly Either section e e or / / may be used, and each 
shows the center of moments to be at U a . The equation is : 

+ 15 833 X 50 - 6 333 X 30 - 6 333 X 10 - L 2 L 3 X 20 = 0; 


L 2 L 3 = +26917. 

The center of moments for UJJ 2 is at L t ; for U 2 U 3 , it is at L 2 ; and 
for U a U t , it is at L y The following equations can now be written : 

+ 20 X U,U 2 + 20 X 15 833 = 6; whence U 1 U 2 = -15 833; 
+ 20 X U 2 U 3 + 40 X 15 833 - 20 X 6 333 = 0; whence U 2 U' 3 = -25 333; 
+ 20 X U,U 4 + 60 X 15 833 - 40 X 6 333 - 20 X 6333 = 0; whence U,U t 
= -28500. 

A diagram of half of the truss should now be made, and all the 
stresses placed upon it. The dimensions should also be put upon this 
diagram. The student should cultivate this habit, as it shows him 
at a glance the general relation of stresses and the general rules of their 
variations. Fig. 33 gives the half-truss, together with the stresses 
and dimensions. The stresses in the members of the right half of 
the truss are the same as those in the corresponding members of the 
left half. 




From inspection of the above diagram, it is seen that the chord 
stresses increase from the end toward the center; that the web stresses 
decrease from the end toward the center; and that all members slant- 
ing the same way as the end-post L U 1 have stresses of that sign, 

Fig. 33. Dimensions and Stress Diagram of Half a Six-Panel Warren Through Truss. 

while all that slant a different way have an opposite sign. These 
relations are true of all trusses with parallel chords and simple systems 
of webbings. 

34. Position of Live Load for Maximum Positive and Negative 
Shears. The dead load, by reason of its nature, is an unchangeable 
load. The stresses due to it are the same at any and at all times. 

* e + y ) 

* y > 


W Ibs. per lin. Foot. 





Fig. 34. Calculating Maximum Positive and Negative Shears in Simple Beam under Live 
Load. Conventional Method. 

With the live load, the case is different. The live load represents 
the movement of traffic upon the bridge. At certain times there may 
be none on the bridge, while at other times it may fill the bridge 
partially or entirely. In such cases the shears due to live load will vary. 


Conventional Method. It has been found that the maximum 
positive shear at any section of a simple beam occurs when the beam is 
loaded from that section to the right support, and that the maximum 
negative shear occurs at the same section when this beam is loaded from 
the section to the left support. This can be proved as follows : 

Let a beam be as in Fig. 34, and let a a be the section under 
consideration. The reaction 7? t is due to the load wy on the part y, 
and to the load wx on the part x. That is, 

Now the shear at the section a a is R t wx; or, 


w xl 2 

wx I 2 ' \-wx\ + ^ = F a . a 
From inspection of this last equation, it is seen that wx 

is the amount that is added to the reaction by loading the part x. 

Also., that T is less than unity, is evident. The amount in 

brackets in the last equation represents the effect of the loading of 
the segment x of the beam. As this is negative and will only reduce 

the positive valued term -~- , it is therefore proved that to get the 

largest positive shear the beam should be loaded from the section to 
the right support. 

From further inspection of the equation, it will be seen that the 
term in brackets, which represents the effect of the load on the seg- 
ment x on the shear, is always negative; and that the term ,which 

represents the effect of the load on the segment y on the shear, is 
always positive. Hence, to get the largest negative shear at the 
section, the load should be on the segment x. That is, the loading 
should be from the section to the left support. 




In a truss, the loads are placed at the panel points; and the 
above rules in application, should be formulated as follows: 

To get the maximum positive shear at a section or in a panel, load all 
panel points to the right of it. 

To get the maximum negative shear at a section or in a pan-el, load all 
panel points to the left of it. 

Example. Determine the maximum positive and the maximum 
negative shears in the panels of the 7-panel Pratt truss shown in Fig. 35, the 




I e 3 

Tat 0= 

Fig. 35. Calculation of Shears in Panels of 7-Panel Pratt Truss. 

live panel load being 40 000 pounds. (It will be noticed that the height of 
the truss is not required.) 

For maximum + V in 1st panel, load L,, L 2 , L 3 , L 4 , L 5 and L e . 

+ V " 2d ' < L 2 , L 3 , L 4 , L 5) and L 6 . 

+ V " 3d ' ' L 3 , L 4 , L 6 , andL 6 . 

4-V " 4th ' ' L 4 , L 5 , and L 6 . 

+ V " 5th ' ' L s andL 6 

+ V " 6th ' ' L 6 . 

+ V " 7th ' ' no panel points at all. 

The reaction produced by each of the loadings is equal to the 
shear for that particular case, since the shear at any section or in any 
panel is equal to the reaction minus the loads to the left of the section 
or panel, and, according to the method of loading, there are no loads 
to the left of the section ; therefore the reaction is equal to the shear. 

For the first panel, the computation is made as follows, the 
center of moments being, of course, at L 7 : 

(fl, = + V'O X 7 X 20 = 40 X 20 + 40 X 2 X 20 + 40 X 3 X 20 + 40 X 
4 X 20 + 40 X 5 X 20 + 40 X 6 X 20. 

It will be seen that as 20 occurs in all terms of this equation, it 
can be factored out by dividing both sides by 20, and the result will 
be the same. The equation can now be written: 



+ F t X 7 = 40 + 40 X 2 + 40 X 3 + 40 X 4 + 40 X 5 + 40 X 6, 
and can still be simplified by writing: 

+ F, -y- (1+2+3 + 4 + 5+6)=+ 120.00, 

which is the form customarily used, the panel length being taken as 
a unit of measurement. The other shears are now easily computed 
in a similar manner: 

+ F 2 = ~ (1 + 2 + 3 + 4+ 5) =+85.71 
+ F 3 = ^-(l+2 + 3 + 4) = +57.14 
+ V 4 = 1 (1 + 2 + 3) = +34.28 
+ V S = 4 (1 + 2) = +17.14 
+ F 8 = 4 (1) = +5.71 
+ F 7 =^-(0) = +0 

In computing the maximum negative shears, sometimes called 
the minimum shears, the reaction is not the same as the shear, as 
there are loads to the left of the section, and these must be sub- 
tracted. The loadings are : 

For maximum V in 1st panel, load no points. 

-V " 2d " " L t . 

-V " '3d " " L ia ndL 2 . 

-V " 4th " " L If L 2 , andL 3 . 

- V " 5th " " L u L 2 , L 3 , and L 4 . 

- V " 6th " " L u L.,, L 3 , L 4 , and L 5 . 

- V " 7th " " L,, L,, L 3 , L 4 , L 5 , and L . 

It is evident that the maximum V 1 is equal to zero, there 
being no loads on the span. The maximum negative shear in the 
second panel is equal to the reaction produced by loading the panel 
point L v and the load at L r Thus, 
77?, = 40 X 6 

- F 2 = /?, - load at L, 




The other shears are next computed as follows: 

_ V 3 = ~ (6 + 5) - 2 X 40 = - 17.14 . 

- V 4 = (6 + 5 + 4) - 3 X 40 = -34.28 


5 + 4 + 3)-4X40 = -57.14 
5 + 4 + 3 + 2) - 5 X 40= -85.71 
- F 7 = -- (6 + 5 + 4 + 3 + 2+ 1)-6X 40 = - 120.00 

The maximum positive and the maximum negative live-load 
shears should now be written side by side, and inspected, in order to 
observe any existing relations which might help to lessen the labor 
of future computations. The values are given in thousands of pounds 





+ 120.00 

- 0.00 

v t 

+ 85.71 

- 5.71 


+ 57.14 



+ 34.28 



+ 17.14 



+ 5.71 


V 7 

+ 0.00 


It is at once seen that the negative shears are numerically equal in 
value to the positive ones, but that they occur in reverse order. This 
simplifies the labor required in the derivation of the negative shears; 
for, after computing the maximum positive shears, these may be 
written in reverse order, and the negative sign prefixed; the result 
will be the maximum negative shears. 

The above method for maximum live-load shears is called the 
conventional method. It is the one that is almost universally used, 
and its use will be continued throughout this text. 

Exact Method. On account of the fact that the floor stringers 
or joists transfer the loads to the panel points, it would be impossible 
to have a full panel live load at one panel point and no load at the 
panel point ahead or behind. In order to have a full panel load at 
one point, the stringers in the panels, on both sides of the point must 




be full-loaded, and this would give a load at the panel point ahead, 
provided the bridge was fully loaded up to and not beyond the panel 
point ahead, equal in value to one-half of a full panel load (see Fig. 




Fig. 36. Illustrating "Exact" Method of Calculating Live-Load Shears in Panels. 

36). The uniform live load, in order to produce full panel loads at 
L 2 , L 3 and L v will also produce one-half a panel load at L v 

By the methods of differential calculus, it can be proved that 
the true maximum positive live-load shear occurs in a panel when the 

m. panels - 

Fig. 37. Calculating Maximum Positive Live-Load Shear in Panel. 

live load extends from the panel point to the right into that panel 
an amount (see Fig. 37) equal to 

in which, 

n = Number of the panel point to the left of the panel under considera- 
tion, counting from the right; 
m = Total number of panels in the bridge; 
p = Panel length. 

Let the truss of Fig. 35 be considered, the live load being 2 000 



pounds per linear foot of truss, and let it be required to determine the 
true maximum positive live-load shear in the 5th panel from the right 




X 20 = 13.333 feet. 

There will now be (4 X 20 + 13.333) X 2 000 = 186 666 pounds on 
the truss; and the left reaction will be { 186 666 X (4 X 20+13.333) 
-i- 2} -=-140 = 62200 pounds. From this must be subtracted the 
amount of the load on the 13.333 feet, which is transferred to the 
point L r This is equal to the reaction of a beam of a span equal to 
the panel length, loaded for a distance of 13.333 feet from the right 
support with a uniform load of 2 000 pounds per linear foot. This 

1 ^ ^^^ 
amounts to (13.333 X 2 OOOX i|) - 20 - 8 890 pounds. The 

true shear is now: 

+ V a = +62 200 - 8 890 = +53 310 pounds. 

The + F 3 , as computed by the conventional method, was + 57 140, 
making a difference of 3 730 pounds between the two. If the true 
shears were computed and compared with the conventional shears, 
it would be found that the V l would be the same, and that the 
remainder of the conventional shears would be greater than the 
corresponding true shears. The difference between any two corre- 
sponding shears would increase from the left to the right end; that 
is, the difference between the conventional and exact shears would 
be greatest in the panel Z- 5 Z> 6 . 

To get the maximum negative shear in any panel, load from the 
left support and out into the panel under consideration an amount 
p x, and proceed in a manner similar to that above described. 

As this method of exact or true shears is seldom employed, 
problems illustrating its application will here be omitted. 

35. Position of Live Load for Maximum Moments. In order 
to obtain the maximum moment at any point, the live load must cover 
the entire bridge. Let the beam of Fig. 34 be considered, and let 
it be required to obtain the maximum moment at the section a a. 
The reaction, as before computed, is: 


all terms of which are positive. The moment at the section is : 


M = R l x x - wx -jr-; 

and substituting for R^ its value, 

wx 2 ( x \ wx 2 wy"x 

-r(^ + y )-^r h ^r ; 
W3 *(JL + JL_ l \ + y 2 . 

X \2l + I ~2j + 21 

But y = I or; therefore, 

The first term of this equation represents the effect of the load 
on the portion x, and the second term represents the effect of the 
load on the portion y. The value of M will always be positive. The 
quantity x varies between and /. When x = 0, M is equal 
to 0. When x = I, the moment is equal to + wy 2 x + 2. For 
all values of x between and /, the first term is positive; and the 
second term being positive in all cases, it is therefore proved that for 
maximum live-load moments at any point, the entire span should 
be loaded, as loads on both segments add positive values to the 
moment value. 

36. Warren Truss under Live Load. In order to analyze a 
truss intelligently, it is necessary to know its physical structure; 
that is, it must be known what character of stress can be withstood 
by the different members. The top chords of all trusses are built 
to take only compression, and the bottom chords are built to take 
only tension; while some web members of some trusses are built for 
tension stresses, some for compression stresses, and some for both. 
The characteristic of the Warren truss is that the web members are 
built so as to be able to withstand either tension or compression. 

Let it be required to determine the live-load stresses in the 
Warren truss of Fig. 32. Let the live load per square foot of roadway, 
which is assumed to be 15 feet wide, be 100 pounds. The live panel 
load is then 100 X 15 X 20 -i- 2 = 15 000 pounds, and the live-load 
reaction under full load is 2^ X 15 000 = 37 500 pounds. 



As the live load must cover the entire bridge to give maximum 
moments and therefore maximum chord stresses, as the <:hord stress 
is equal to the moment divided by the height of the truss a simple 
method for the determination of live-load chord stresses presents 
itself. The live load and the dead load being applied at the same 
points, and being different in intensity, the stresses produced will 
be proportional to the panel loads. The maximum live-load chord 
stresses (see Fig. 33) will then be equal to the dead-load chord stresses 
multiplied by 15 000 -+ 6 333 = 2.371, and they are as follows: 

Lot/! = -2.371 X 17700 = -42000 
U 1 U 2 = -2.371 X 15833 = -37530 
U 2 U 3 = -2.371 X 25333 = -60050 
U 3 U 4 = -2.371 X 28 500 = -67 600 
L L, = +2.371 X 7917= +18770 
L,L 2 = +2.371 X 20583 = +48800 
L 2 L 3 = +2.371 X 26917 = +63850 

The next step in order is to determine the maximum positive 
shears, and from these write the maximum negative shears. This 
is done as follows: 

+ Live-Load V Live-Load V 

(1 + 2 + 3 + 4+ 5) = +37 500 

V 2 = (i + 2 + 3 + 4) =+25000 -2500 

F 15_000 + 2 + 3) =+15000 -7500 

= 15000 + 2) = + 7 500 - 15 000 

= + 2500 -25000 

y 6 = +0 -37500 

The stresses produced by the positive shears are called the 
maximum live-load stresses, and are : 

+ L U l cos < + 37 500 = .-. L U l = -37 500 X 1.12 = -42 000 

- U,L, cos + 37 500 = .'. U,L, = +37 500 X 1 12 = +42 000 
+ L 1 U, cos + 25 000 = .-. L,U 2 = -25 000 X 1.12 = -28 000 

- U 2 L 2 cos <j> + 25 000 = .-. U 2 L 2 = +25 000 X 1.12 = +28 000 
+ L 2 C7 3 cos < + 15000 = .-. L 2 C7 3 = -15000 X 1.12 = -16800 

- U 3 L 3 cos <j) + 15 000 = .-. U 3 L 3 = + 15 000 X 1.12 = +16 800 

The stresses produced by the negative shears are called the 
minimum live-load stresses, and are: 




+ L C7, cos <; 

!> + =0 

.-. L U l = 

U l L l cos (j 

!> + =0 

.-. t/'.L, = 

+ L! f/ 2 COS <; 

& - 2 500 = 

.-. L^ a = +2500 X 1.12 = 

+ 2800 

- [7 2 L 2 cos <; 

& - 2 500 = 

.-. U 2 L 2 = -2500 X 1.12 = 


+ L 2 t/ 3 cos c 

b - 7 500 = 

.-. L 2 U 3 = +7500 X 1.12 = 

+ 8400 

- U 3 L 3 cos ^ 

6 - 7 500 = 

.-. U 3 L 3 = - 7 500 X 1.12 = 


These stresses, together with the dead-load stresses, should 
now be placed together as a half-diagram, as is done in Fig. 38, the 
stresses being rounded off to the nearest ten pounds and then ex- 
pressed in thousands of pounds. No minimum live-load stress is 
given for the chords, as this will evidently be zero in all cases, since 
no position of the live load will cause a reversal of stress. It will be 
seen that the stresses produced by the negative shears are of opposite 

dl - 1583 
U,MU - 37.53 

28 50 

Pig, 38. Dimension and Stress Diagram of Warren Half-Truss under Live Load. 

sign from the stress produced by the dead load, and these tend to 
decrease the dead-load stress by that amount; and in some cases 
(see L 2 U 3 and U 3 L 3 , Fig. 38) it will be so large as to overcome the 
dead-load stress and therefore change the total stress from one kind 
to another. Do not forget, in considering any combination of the 
above stresses, that the dead load occurs with either the maximum 
or the minimum live load, but not with both at the same time. 

37. Counters. By reference to U 3 L 3 (Fig. 38), it is seen that 
when the live load is on the panel points L l and L 2 the total stress in 
the member is + 3.54 + ( 8.40) = 4.86, a compressive stress of 
4 860 pounds ; whereas, under dead load alone, the stress was + 3.54, 
a tensile stress of 3 540 pounds. If the member U 3 L 3 had been built 
of long, thin bars which could take only tension, and which con- 
sequently would have doubled up under the resultant compression 



brought upon them by the combined 
action of the dead and minimum live- 
load stresses, then this member could 
not be used in this case, but some 
other arrangement would be necessary 
in order to insure the stability of the 

In the Warren truss, no special ar- 
rangement is necessary, as the web 
members are built so as to take either 
tension or compression; but with the 
Pratt and Howe trusses some special 
arrangement is necessary, as the diag- 
onals are built to take one kind of 
stress only. The case of the Pratt will 
be considered first. 

The Pratt truss has the diagonals 
made of long bars which take tension 
only, and the intermediate posts are 
constructed so as to be able to take 
compression only. It is not necessary 
to consider the intermediate posts, for 
the action of the web members is such 
that the resulting stresses are always 

Let the 13-panel Pratt truss of 
Fig. 39 be considered. The panel 
length is 18 feet, the height 25 feet, the 
dead panel load 22 000 pounds, and 
the live panel load 58 500 pounds. 
The secant is (18 2 + 25 2 )* -=- 25 = 
1.231. The dead-load shears and the 
maximum and minimum live-load 
shears are placed directly below their 
respective panels. Only those mem- 
bers are shown full-lined in Fig. 39 
which act under the dead load. Note 
that the dead -load shears in the center 



panel being zero, "the dead-load stress in the diagonals in the center 
panel would be X sec <f> = 0. 

In the first four panels from either end, the live-load shear, 
which is of a different sign from that of the dead-load shear, is 
smaller than the dead -load shear, and therefore will not cause a 
reversal of stress in the member in that panel. For example, take 
U 3 L 4 ; then, for dead-load stress, 

-C/ 3 L 4 cos + 66.0 = .:U,L a = +66.0 X 1.231 = +81.20 

For live-load stress, 

-7 3 L 4 cos <f> - 27.0 = .-.U a L, = -27.0 X 1.231 = -33.25 

The total stress = + 81.20 - 33.25 = + 47.95, which is still 

Considering L g U 10 , the stress equations are : 
For dead-load stress, 

+ L 9 C7 10 cos (j) - 66.0 = .'. L 9 C7 10 = +81.20 

For live-load stress, 

+ L u [/ 10 cos <j> + 27.0 = L U M = -33.25 

The total stress, as before, is + 47.95, or a tension of 47 950 pounds. 
An inspection of the center panel and the two panels on each 
side of it, shows that the live-load shear is of a different sign from 
the dead-load shear, and is also greater in value than the dead-load 
shear. If the members shown in Fig. 39 were the only ones in the 
panels, then the dead-load stresses would be: 

- U t L s cos $ + 44.0 = U 4 L S = +54.20 

- C7 5 L 6 cos <j> + 22.0 = f/ 5 L 6 = +27.10 
+ L 7 [/ S cos - 22.0 = L 7 U S = +27.10 
+ L 8 i7,eofl i - 44.0 = L 8 U 9 = +54.20 

and the live-load stresses caused by the shear of opposite sign from 
that of the dead-load shear, are: 

-t/ 4 L 5 cos - 45.0 = U 4 L, = -55.40 

- U 6 L K cos - 67.5 = t/ 8 L 6 = -83.10 
+ L 7 U s cos <f> + 67.5 = L 7 U S = -83.10 
+ L s [/ 9 cos <f> + 45.0 = L s U a = -55.40 

As no diagonal acts under dead load in the center panel, we may 
assume that U 6 L 7 acts under live load. The stresses which occur 

in this are: 

+ U,,L 7 cos $ + 94.5 = U K L 7 = +116.30 

- U b L 7 cos - 94.5 = C7 8 L 7 = - 116.30 



The above shows that compressive stresses will occur in the 
diagonals which were built for tension only. These stresses are : 

U 4 L 5 = +54.20 - 55.40= - 1 200 pounds 

U & L 6 = +27.10 - 83.10= -56000 " 

L 7 U S = +27.10 - 83.10= -56000 " 

L s U g = +54.20 - 55.40 = - 1 200 " 

U 6 L 7 = - 116.30 = -116300 " 

If some provision were not made for these stresses, they would cause 
the members to crumple up and the truss to fail. In order to allow 
for them, diagonals are placed in the panels, as shown by the dotted 
and dashed lines. These members will take up the above stress; 
and moreover, as they slope the opposite way from the main members, 
they will be in tension. 

In order to prove this, assume L b U 6 to act when the live load is on 
points L 5 , L 4 , L y L 2 , and L v Now, U 5 L 6 will not be regarded, as its 
stress will be zero. Then the stresses will be : 
For dead load, 

+ L 5 [7 6 cos < + 22.0 = L & U 6 = -27.10. 

For live load, 

+ L s t/ 6 cos - 67.5 = L 5 C7 6 = +83 10; 

and the total stress in L 5 *7 6 will be - 27.10 + 83.10 = + 56.00. 

In a similar manner, the stresses in the other members are:LJJ 5 
= +1.2; L 6 U 7 =+ 116.30; U 7 L S =+ 56.00; and UgL 9 = + 1.2. 
These diagonals are called counters or counter-bracing. 

From a consideration of the foregoing, it is evident that: 

(a) // the live-load shear in any panel is of opposite sign and greater 
than the dead-load shear in the same panel, then a counter is required. 

(5) The stress in a counter is equal to the algebraic sum of the dead-load 
shear and the live-load shear of opposite sign times the secant of the angle it 
makes with the vertical. 

This is true for any truss with horizontal chords and a simple system 
of webbing with diagonals and verticals. 

38. Maximum and Minimum Stresses. Some specifications 
require the member to be designed for the maximum stress, while 
others take into account the range of stress. In this latter case it 
is necessary to determine the minimum as well as the maximum stress. 
Except where a reversal of stress occurs and this does not happen 
in trusses with horizontal chords few specifications require any 




but the maximum stresses to be com- 
puted. For that reason, little space 
will here be devoted to the minimum 
stresses, their computation in succeed- 
ing articles being thought to illustrate 
them sufficiently. 

(a) The maximum stress in a member 
is equal to the sum of the dead-load stress 
and the live-load stress of the same sign. 

(b) The minimum stress is equal to 
the sum of the dead-load stress and the live- 
load stress of the opposite sign, or to the dead- 
load stress alone, according to which gives 
the smallest value algebraically. By this 
latter statement it should be seen that if 
the maximum stress is 58.60, then or 
+ 18.00 would be smaller than -3.00. 

(c) It is evident that the minimum 
in all counters and in all main members 
in panels where counters are employed will 
be zero, for when the counter is acting the 
main member is not, and therefore its stress 
is zero. The reverse is also true. 

(rf) An exception to a is seen in the 
case of the counters. Here it is evident 
that the maximum stress is equal to the 
algebraic sum of the dead-load shear and 
the live-load shear of opposite sign times 
the secant of the angle which the counter 
makes with the vertical. 

While it is true that in trusses with 
horizontal chords the loading for maxi- 
mum shears will give the maximum 
live-load stress to be added to the 
dead load for the maximum stress, it is 
not always true that the loading for 
minimum live-load shears will give the 
stress to add to the dead-load stress to 
get the minimum stress. However, the 
loading for the minimum live-load 
shears will give the live-load stress to 
be added to the dead-load stress for 
the minimum stress, except in the case 




of verticals placed between panels each of which contains counters, 
and in that case it may or may not do so. In such cases a loading 
must be assumed preferably the one for minimum shears and 
the shears in the panels on each side of the vertical must be com- 
puted for the loading assumed. 
If the resultant shear is the same 
sign as the live load, then the 
main diagonal acts; if it is of 
different sign, then the counter 

As an example, let it be re- 
quired to find the minimum stress 
in the vertical U 5 L 5 of the truss 
of Figs. 39 and 40. It is assumed 
that the loading for minimum 
shears will give the result. The 
section a a is then passed, and 
the live load placed on L 5 and all 
points to the left. The shears will 
then be as shown in Fig. 41. To 
obtain the shear in the panel L t L 5 , 

d.l. 22.0 
11 58.5 


4- 44.0 

.+ 22.0 



- 67.5 



Fig. 41. Stress Diagram for Vertical in 
Truss of Fig. 40. 

under this loading, it must be re- 
membered that a load is at L 5 ; and so the shear is the shear in the 
panel L 5 L 6 with the panel load at L 5 added, or, 67.5 + 58.5 = 
9.0. The diagonals now act as indicated by Fig. 41, and the total 
stress in U b L 5 is determined by passing a circular section around t/ 5 , 
and it is : 

-Load at U, - U 5 L 5 = 0. 

As there is no load at U 5 , the stress in U 5 L 5 is = 0. The same result 
will occur if points Z/ 4 or L 3 and to the left are loaded; but if points 
L 2 and to the left are loaded, the members U^L 5 and U b L 6 will act, 
and the stress in U 5 L 5 will then be equal to the shear on the section 
a a. The stresses are: Dead-load, 22.0; and live-load, + 
13.5, which gives a total of 8.5; but as the maximum stress is 
-22.0 - 126.0 = - 148.0, it is evident that and not -8.5 is the 

The computation of the maximum stress is as follows: 

Load points L 6 and to the right. The shear on a a is, for 



dead load, +22.0; and for + live load, +126.0; and the equations of 
the stresses are: 

+ 22.0 + C/ 5 L 5 = U 5 L S = - 22.0 

+ 126.0 + C7 5 L 6 = U b L s = - 126.0 

Max. = -1480 


39. The Pratt Truss. The Pratt truss is used to perhaps a 
greater extent than any other form; probably 90 per cent of all simple 
truss spans are of this kind. 

Let it be desired to determine the stresses in the 8-panel 200-foot 
single-track span shown in Fig. 42, the height being 30 feet, the dead 
panel load being 30 000 pounds, and the live panel load 62 400 

pounds. The secant is ,("25* +~30 2 ) * -i- 30 = 1.302, and the cosine 

is 0.7685. The dead-load reaction is 3 X 30.0 = 105.0. 
The dead-load shears are : 

F, = +105.0 
V, = + 75.0 
V 3 = + 45.0 
V 4 = + 15.0 
V t = - 15.0 

The dead-load chord stresses may be tabulated as follows (see 
Articles 27 and 29): 

Dead- Load Chord Stresses 






Z/ L, = L 1 L 2 

a a 

u t 

+ 105.0 X 25 - 

Z/,L 2 X 30 = 

+ 87.5 

L 2 L 3 


U 2 

+ 105.0 X 50 - 

30.0 X 25 - L 2 L 3 X 

30 = 

+ 150.0 

L 3 L 4 

c c 

U 3 

+ 105.0 X 75 - 

30.0 (25 + 50) - L 3 L 4 

X 30 = 

+ 187.5 

U,U 2 

a a 

L 2 

+ 105.0 X 75 - 

30.0 X 25 + C7,[/ 2 X 

30 = 




L 3 

+ 105.0 X 75 - 30.0 (25 + 50) + 

U 3 U t 

c c 

L 4 

U 2 U 3 X 30 = 
+ 105.6 X 100 - 30.0 (25 + 50 + 75) 


+ U,U 4 X 30 



In determining dead-load stresses in web members, it is cus- 
tomary to assume one-third of the dead panel loads as applied at the 




upper chord points. This, as will be seen, makes no difference in the 
stresses in the chords or in the diagonals, the stresses in the verticals 
only being different from what is the case when all the dead load is 
taken on the lower chord. 

The stresses in the diagonals (see Articles 27,28, and 30) are: 

Dead = Load Stresses in Diagonals 









o o 

+ 105.0 

+ 105.0 

+ L n U 

i X 

0.7685 = 



a a 

+ 75.0 

+ 75.0 

- U } L 

n X 

0.7685 = 

+ 97.60 

U L, 


+ 45.0 

+ 45.0 

- U 2 L 

t X 

0.7685 = 

+ 58.60 

U 3 L 4 

c c 

+ 15.0 

+ 15.0 

- U 3 L 

4 X 

0.7685 = 

+ 19.53 

In determining the stresses in the verticals, it is to be remem- 
bered that one-third the dead panel load (or 10.0) is at the panel 

o a 'i b 's'c 3' 

Fig. 42. Outline Diagram of 8-Panel Single-Track Pratt Truss Span. 

points of the upper chord, and two-thirds (or 20.0) is at the lower 
chord. The stress in the hip vertical U l L l is determined by passing 
a circular section around L,. It is solved thus : 

-20.0 + U^L, = U^ = +20.0 
In a similar manner the stress in U 4 L 4 is found to be : 

-10 - t/ 4 L 4 ' = C7 4 L 4 = -10.0 

In order to find the stress in the remaining verticals, sections 1 1 
and 2 2 are passed, cutting them, and the shears on these sections 
computed. The shears are: 

F t - , - + 105.0 - 2 X 20 - 1 X 10 = + 55.0 
F 2 _ 2 = + 105.0 -3X20-2X10= +25.0 

The stress equations are written, remembering that as the verticals 




U i 

make an angle of zero with the vertical, their cosine is equal to unity. 

These equations are: 

+ U 2 L 2 + 55.0 = U 2 L 2 = -55.0 

+ U 3 L 3 + 25.0 = U 3 L 3 = -25.0 

The live-load chord stresses will be proportional to the dead- 
load chord stresses, as both loads cover the entire truss in exactly 
the same manner. The ratio of the panel loads by which the dead- 
load chord stresses are multiplied in order to get the live-load chord 

stresses, is ^7^ = 2.08, and the chord stresses are : 
oO 000 

L a L, = L,L 2 = + 87.5 X 2.08 = +182.0 

L 2 L 3 = +150.0 X 2.08 = +312.0 

L 3 L 4 = +187.5 X 2.08 = +390.0 

VJJ* = -150.0 X 2.08 = 312.0 

U 2 U a = -1875 X 2.08 = -390.0 

U 3 U 4 = -200.0 X 2.08 = -416.0 

As the entire bridge is to be loaded to get the maximum stress in 
L U 19 it is therefore equal to the 
dead-load stress times the above 
ratio; or LJJ l = -136.70 X 2.08 
- 284.20. 

The maximum live-load stress 
in U^ is determined by passing 
a circular section around L v and 
is solved (see Fig. 43) from the 
equation : 

+ U,L - 62.4 = /. U.L, = +62.4 
For UJj v the section a a is 
passed, and the points L 2 and to 
the right are loaded. The maxi- . 

mum shear is: 

+ V 2 = + -^ (1+2 + 3 + 4 + 5 + 6)= + 163.8; 


and the stress equation is : 

+ 163.8 - C/iL 2 X 0.7685 - .'. E7,L 2 = +213.2. 

In a similar manner, pass section 6 6, and load points L 3 and to 
the right, and the shear and the stress equations for U 2 L a are: 

co 4 

+ F 3 = + (1 + 2 + 3 + 4 + 5)= +117.0 

+ 117.0 - U 2 L 3 X 0.7685 = .'. U 2 L 3 = +152.4 





For UJjv the section c c is passed, and the panel points to the 
right are loaded. The shear and stress equations are : 

+ V 4 = + ~ (1 + 2 + 3 + 4)=+ 78.0 


+ 78.0 - C7 3 L 4 X 0.7685 = /. U 3 L t = +101.6 

For the maximum stresses in the verticals, sections 1 1, 2 2, 
and 3 3 are passed, and in each case the panel points to the left 
of these loaded. The shears are 

(1+2 + 3 + 4 + 5)= +117.0 


V 2 -, = ^- (1 + 2 + 3 + 4) = +78.0 

V,- 3= -^(1 + 2 + 3) = +46.8 

The stress equations for U^L 2 and U 3 L 3 are simple, as only three 
members are cut. They are : 

+ 117.0 + U 2 L 2 = .-. / 2 L 2 = - 117.0 

+ 78.0 + U 3 L 3 = /. U 3 L 3 = - 78.0 

It is seen that the section 33 cuts the member LJJ 5 , and 
therefore the stress in this must 


\ L, 

be determined before the stress 
equation can be written, as its ver- 
tical component will enter into it. 
However, by comparing the dead- 
load shear in that panel, which is 
15.0, and the live-load shear V 3 _ 3 , 
which is +46.8, it is seen that the 
resultant shear is + ; and, as this 
is of opposite sign from the dead- 
load shear, a counter is required 
and is acting. The stress in 7 5 L 4 is zero, and the diagonals act as 
in Fig. 44, the section 3 3 then cutting three members. The 
stress equation is + 46.8 + U t L t = 0, from which U,L 4 - - 46.8. 
Care should be taken not to add to this -46.8 the -10.0 
derived as dead-load stress on page 44, in order to get the maximum 
stress, as the 10.0 previously derived was the dead-load stress 
in U 4 L 4 when U 3 L 4 and LJJ 5 were acting. The dead-load stress which 
goes with the live-load stress of 46.8 acts simultaneously with it, 

Pig. 44. Calculation of Stress in Diag- 
onal of Span of Fig. 42. 




and is the dead-load stress in U t L 4 when the members U,L 4 and U 4 L 5 
are acting as in Fig. 44. The dead-load shear on the section 3 3 
would then be the left reaction minus the loads at points U v U v U 3 , 
L I} L 2 , L 3 and Z/ 4 ; or, 

F 3 _ 3 = +105.0 -3X10-4X20= -5.0; 

-5.0 + U t L 4 = .-.U 4 L 4 = +5.0. 

Remember that this + 5.0 can act only when the live load tends to 
produce a stress of 46.8; and thus the total stress in U 4 L t with live 
load in that position is -46.8 + 5.0 = -43.8, while with dead load 
only in the truss the stress is 10.0. 

The dead-load shears and the maximum + and live-load 
shears should now be written for inspection, in order to investigate 
for counters and then for the minimum stresses. Those whose 
derivation has hot been given should be easily computed by the stu- 
dent at this time. The shears are 




F, + 105.0 
V 2 + 75.0 
V 3 + 45.0 
V t + 15.0 

+ 218.4 
+ 163.8 
+ 117.0 
+ 78.0 

- 7.8 

From a study of these it is seen that a counter is required in the 
4th panel according to rule a, Article 37; and according to rule b of 
the same article, the maximum stress is ( 46.8 + 15.0) X 1.302 = 
+ 41.4, the minimum stress for it and also U 3 L 4 being zero according 
to the same article. A counter is also required in panel 5, as the 
truss is symmetrical. 

The minimum live-load stress in U l L l is zero, and occurs when 
no live load is at the point L r 

The minimum live-load stresses in the diagonals UJL 2 and U 2 L 3 
occur when the truss is loaded successively to the left of the sections 
a a and b b, in which case the shears are 7.8 and 23.4 
respectively. The stress equations are 

- U,L 2 - 7.8 X 0.7685 = .'. f/,L 2 = - 10.16 

-U 2 L 3 - 23.4 X 0.7685 = .'. U 2 L 3 = -29.15 

The minimum live-load stress in UJL 2 is obtained by passing 




section 1 1 and loading the panel points to the left. The live-load 
shear is the same at this section as it is at the section 6 6 namely, 
23.4. The stress equation is 

+ U 2 L 2 - 23.4 = .'. U 2 L 2 = +23.4 

To determine the minimum live-load stress in U 3 L 3 , proceed as 
indicated on page 42. By loading points L 3 and to the left, the live- 
load shear in the 4th panel will be 46.8, and in the 3d panel under 
this same loading it will be -46.8 + 62.4 = + 15.6. The sign of the 
total shear in the two adjacent panels, and the members acting, are 
shown in Fig. 45. The stress in U 3 L 3 is then determined by using a 



4- 10 
/ 4U 5 








+ 4-5.O 

+ I5.O 


+ 1 5.6 




+ 15.0 

- 15.0 


- 15.6 


Total v 

Fig. 45. Fig. 46. 

Stress Diagrams for Verticals in Span of Fig. 42. 

circular section around U 3 , and is simply the dead load at U 3 , there 
being no live-load stress in the member when the bridge is loaded as 
has been done. 

In finding the minimum live-load stress and also the minimum 
stress in U t L 4 , the same method of procedure will be followed. Let 
LI and to the left be loaded. Then the shear in the 5th panel is 
78.0, and under this same loading the shear in the 4th panel is 
-78.0 + 62.4 = -15.6. The sign of the total shear in each of the 
adjacent panels is given in Fig. 46. It should be remembered that 
a resultant shear with the same sign as the dead -load shear causes 
the main diagonal to act, while a resultant shear of opposite sign to 
that of the dead-load shear causes the counter to act. The members 






acting are shown, and a section 4 4 can be passed. The dead-load 
shear at this section is 105 - 3 X 20 - 4 X 10 = +5.0; and 

- U 4 L 4 + 5.0 = 0. 

UJj t = +5.0 = Dead-load stress in this case. 
The live-load stress which acts at the same time is: 

-U 4 L 4 - 15.6 = .-. U 4 L 4 = -15.6, 

the term 15.6 representing the live-load shear on the section 4 4. 
This is not the minimum stress, as will next be shown, but it illus- 
trates the fact that the loading for minimum live-load shears does 
not always give the minimum 
live-load stress. 

By loading L v the live-load 
shear in the second panel, and 
likewise all others from this to 
the right support, will be 7.8. 
The total shears, together with 
their sign, and also the members 
they cause to act, are given in 
Fig. 47. The minimum live-load 
stress in U 4 L 4 is found to be zero, 
and the dead-load stress is 10, 
as is derived by passing a circular 
section around U 4 , the equation 








+ 15.0 



- 7.8 

- 7.8 

Total v 


Fig. 47. Stress Diagram for Vertical in 
Span of Fig. 42. 

being as follows: 

-Live load at U 4 - U 4 L 4 = .'. U 4 L 4 = for live load. 
-Dead load at U 4 - U 4 L 4 = .'. U 4 L 4 = -10.0 for dead load. 

A diagram of half the truss should now be made, and all dead 
and live load stresses placed upon it, and these should be combined so 
as to form the maximum and the minimum stresses. Such a dia- 
gram, together with all stresses, is given in Fig. 48. 

The stresses are written in the following order: Dead load, 
maximum live load, minimum live load, the maximum, and the 
minimum. In the chord and end-post stresses, there is no minimum 
live-load stress recorded, it being zero. Where pairs of stresses occur 
simultaneously, a bent arrow connects them. 

40. The Howe Truss. The physical make-up of the Howe truss 




differs from that of 
the Pratt in that 
the diagonals are 
made to stand com- 
pression only, and 
the verticals can 
stand tension only. 
In the Pratt truss 
it was found that 
none of the inter- 
mediate posts could 
be brought into 
tension by any 
loading. In the 
Howe truss it will 
be found that none 
of the verticals can 
be brought into 

Let it be required 
to determine the 
stresses in a Howe 
truss of the same 
span, height, and 
loading as the Pratt 
truss of Article 39. 
An outline diagram 
is given in Fig. 49. 

The dead-load 
shears and the 
maximum and min- 
imum live-load 
shears will be the 
same as for the 
Pratt truss, and 
they are : . 







V, + 105.0 

+ 218.4 


V, + 75.0 

+ 163.8 

- 7.8 

V, + 45.0 

+ 117.0 


V 4 + 15.0 

+ 78.0 


K 5 - 15.0 

+ 46.8 


Inspection of these shows that counters are required in the 4th 
and 5th panels (see Article 37). 

The dead-load lower chord stresses will be computed by the 

U 7 

Fig. 49. Outline Diagram of 8-Panel Single-Track Howe Truss Span. 

tangent method (see Article 31), the section being y y lf etc. The 
tangent of <f> is 25 -+ 30 = 0.8333. The stresses may be conven- 
iently tabulated as follows: 

Dead-Load Chord Stresses (Lower Chord) 







_ V| 

- 105.0 X 0.8333 + L L, = 

+ 87.5 

L,L ? 


- 2/2 

-(105.0 + 75.0) 0.8333 + L,L 2 = 

+ 150.0 


- 2/3 

-(105.0 + 75.0 + 45) 0.8333 + L 2 L, = 

+ 187.5 

L 3 L 4 


- 2/4 

-(105.0 + 75.0 +45.0 +15.0) 0.8333 + L 3 L 4 = 

+ 200.0 

A simple method for the determination of the upper chord 
stresses, is to pass a section and to equate the sum of the horizontal 
forces. Pass section 1 1. The only horizontal forces are the 
stresses in L L l and UJJ^; and as these are parallel, one must be equal 
and opposite to the other. In a like manner the stresses in the other 
sections of the top chord are found. The stresses are : 
U,U 2 = -L L, = -(+ 87.5) = - 87.5 
U 2 U 3 = -L,L 2 = -( + 150.6) = -150.0 
U 3 U 4 = -L 2 L 3 = -( + 187.5) = -187.5 



A consideration of the Pratt truss shows that this method can be 
applied to it in determining the chord stresses. 

As it is known that the diagonal web members are in compression 
under the dead load which produces a positive shear in the left half 
of the truss, it is evident that positive live-load shears will produce 
compressive stresses, and negative live-load shears tensile stresses, 
in the diagonals in the left half of the truss. Also, from Article 30, 
the stress in a diagonal is V sec <. The stresses can now be written 
directly without the aid of the stress equation : 

L U l = -105.0 X 1.302 = -136.70 

L,C7 2 = - 75.0 X 1.302 = - 97.60 

L 2 U 3 = - 45.0 X 1.302 = - 58.60 

L a U t = - 15.0 X 1.302 = - 19.53 

Likewise the stresses in the verticals can be written directly, remem- 
bering that here the secant is unity, and that the shear at the section 
cutting the member is to be used, not forgetting that ^ of the dead 
panel load is applied at the top panel points. The shears and 
stresses are: 

V. - , = +105.0 - 10 = +95.0 Z7,L, = +95.0 

Fo_ 2 = +105.0 - 20 - 2 X 10 = +65.0 U. 2 L 2 = +65.0 

V 3 _ 3 = +105.0 -2X20-3X10= +35.0 U 3 L X = +35.0 

The member U^L 4 cannot be easily determined by passing a 
section 44, for this cuts four members. It is determined by passing 
a circular section about the point L 4 , the equation being + U 4 L 4 
- 20.0 = 0, from which C7 4 L 4 = + 20.0, which is equal to the dead 
panel load at the point Z 4 . 

The live-load chord stresses are determined by multiplying the 
dead-load chord stresses by the ratio of the live to the dead loads. 
This has been found to be equal to 12.08. The live-load chord stresses 
are found to be : 

L L, = +182.0 U,U 2 = -182.0 

L,L 2 = +312.0 U 2 U 3 = -312.0 

L 2 L 3 = +390.0 U 3 U 4 = -390.0 
L 3 L< = +416.0 

As the character of the stresses which can be taken by the 
diagonals and the verticals is known, the maximum and minimum 
live-load stresses can be written without first writing the stress 
equations. The maximum live-load stresses are: 



L U l = -218.4 X 1.302 = -284.36 tf.L, = +218.4 

L 1 U 2 = -163.8 X 1.302 = -213.27 U 2 L 2 = +163.8 

L 2 U 3 = -117.0 X 1.302 = -152.33 U 3 L 3 = +117.0 

L 3 U 4 = - 78.0 X 1.302 = -101.56 f/ 4 L 4 = + 78.0 

It should be noted that when Z/ 4 and all panel points to the right 
are loaded, the shears and the members acting are as shown in Fig. 
50. The dead-load shear on the section 4 4 is +15.0, less the load 
at U 4 , or + 15.0 10.0 = +5.0; and the equation of stress is U 4 L 4 
+ 5.0 = 0, from which J7 4 L 4 = +5.0. Thus it is seen that in this 














+ 15.0 



-- 7.6 

+ 15-6 

Total v 






+ 15.0 




Total v 



Fig. 50. Fig. 51. 

Stress Diagrams for Members of Howe Truss Span of Fig. 49. 

case the dead-load stress is +5.0 when the live-load stress is +78.0. 
The maximum stresses in the counters (see Article 37) are: 

(-46.8 + 15.0) 1.302 = -41.4. 
The minimum live-load stresses are now written as follows: 

L,C7 2 = + 7.8 X 1.302 = +10.16 
L 2 U 3 = +23.4 X 1.302 = +29.15 
L 3 U 4 = 
U 3 L< = 

tf,L, - 

U 2 L 2 =-- -7.8 

U 3 L 3 | See discussion 

f/ 4 L 4 \ following. 

If live panel loads were placed at points L v L v and L 3 the live- 
load shear in c c would be 46.8; and the dead-load shear being 
+ 15, the counter would act, and the stress in UgL a would be tensile 
and equal to the sum of the dead and live panel loads which are at 
its lower end L r If points L } and Z/ 2 had live panel loads on them, 



the resultant shear in c - c would be -23.4 + 15.0 - -8.4; the 
counter would act, and the stress .in U t L 4 would be tensile and equal 
to the dead panel load which is at L y There being no live panel load 
at I/g, the live-load stress in U^ would be zero under this loading. 
If a live panel load be placed at L^ only, then the shears and the mem- 
bers acting will be as shown in Fig. 51, and V 3 3 for dead load = 
+ 45.0 - the load at f/ 3 , or - 45 - 10 - +35.0. The V 3 3 for 
live load = 7.8, and the stress equation U 3 L 3 7.8 = 0, from 
which U 3 L 3 = 7.8. So this live-load compression stress of 7 800 
pounds occurs at the same time as the dead-load tensile stress 
of 45 000 pounds. 

By loading various groups of panel points in succession and 
determining the resulting live-load stresses in U t L 4 , it will be found 
that under no loading can a negative live-load stress be produced. 
The minimum live-load stress is therefore zero, and occurs when 
there is no live load on the bridge. 

The stresses should now be placed on an outline diagram similar 
to that of Fig. 48, and the stresses in corresponding members com- 
pared with those in that figure. This is left for the student. 

41. Bowstring and Parabolic Trusses. A bowstring truss is 
shown in Fig. 13, the full lines representing the main members, which 
are the members under stress by the dead load. The dotted members 
represent counters which may be stressed by the action of the live 

As before mentioned, the stresses in the chords and also in the 
webbing are quite uniform. When the end supports and the panel 
points lie on the arc of a certain curve, called a parabola, then, under 
full load, the stresses in all panels of the lower chord are equal; the 
stress in all verticals is tensile and is equal to the panel load at the 
lower end ; and the stress in all diagonals is zero. Under partial load, 
the stresses in the webbing are exceedingly small, and the chord 
stresses remain almost equal. 

If it is desired to have a parabolic truss, first decide upon the 
length of span, the number of panels, and the height at 
the center. The height of any vertical post is given by the 
formula : 

, rr 

h = H 



in which, 

H = Approximate height at center; 
d = Distance of vertical post from center; 
I = Span; 
h = Height of vertical post sought. 

All distances are in feet. Suppose, as an example, that it was 
desired to determine the heights of the vertical posts in an 8-panel 
parabolic truss of a height approximately equal to 24 feet. One-half 

Fig. 52. One-Half of 8-Panel Parabolic Truss. 

the truss is shown in Fig. 52. At the center, d = 0, and the equation 
reduces to h = H, which is 24 feet. For U 3 L 3 , d = 20; then, 

4 X 24 X 20 2 - 

h = 24 - 

from which, 
For t/ 2 L 2 , 

160 2 
h -= 22.5 feet. 

d = 40 

4 X 24 X 40 2 
160 2 

h = 24 ^-~ 

For U,L V 

d = 60 

= 24 - 4 x 24 x ^ a 
160 2 

= 18.0 feet. 

= 10.5 feet. 

Inspection of the above results shows that the span or the center 
height must become quite great before the clearance at U t L t will be 
sufficient to allow the traffic to pass under a portal bracing at this 
point. For this reason these trusses are usually built as through 
trusses with bracing on the outside of the truss, which connects to 
the floor-beams extended. 




In the bowstring truss, 
the panel points of the 
top chord usually lie on 
the arc of a parabola 
which does not pass 
through the supports. 
For example, suppose 
that it was decided to 
have the span and pan- 
els the same as shown in 
Fig. 52, but the height 
at L l was to be '28 feet, 
and at 4 36 feet. By 
substituting these values 
in the equation just 
given, and solving for /, 
the place will be deter- 
mined where the para- 
bolic curve cuts the 
lower chord extended, 
and the lengths of the 
vertical posts may be 
computed as before. 
Substituting these re- 

28 = 36 - 

4 X 36 X 60 2 

(-36 + 28) l a = - 4 X 36 
X 60 2 

7 _ / 4 X 36 X 6"0" 2 

= 254.5, 

which shows that the 
arc cuts the lower chord 
extended at a point 
254.5 -T- 2 = 127.25 feet 
from the center of the 
span (see Fig. 53). 


The other vertical posts are: 
U,L, k = 30 _*J6J^g_ 3;ul feet . 

= 32.44 feet; 

4 V *3fi V fiO 

[/.L! /i = 36 - = 5 = 28.00 feet, which checks. 


The analysis of a bowstring truss will now be given. Both the 
maximum and minimum stresses will be determined, as reversal of 


Fig. 54. Outline Diagram of 5- Panel Bowstring Truss Span. 

stresses is liable to occur in the intermediate posts. The loading for 
minimum live-load stresses can be ascertained only by trial, care 
being taken to compute the dead-load stresses for the arrangement 
of web members caused by that particular live loading. 

Let it be required to determine the maximum stresses in the 
5-panel 100-foot bowstring truss 
shown in Fig. 54, remembering 
that the diagonals take only ten- 
sion. The height of U^ is 20 
feet, and of U^ 25 feet. The 
dead panel load is 17 200 pounds, 
and the live panel load is 50 000 
pounds. The full lines show the 
main members which act under 
dead-load stress, and the dotted 
lines show the counters which may 
act under the action of the live load. 


Fig. 55. Resolution of Forces around 

Panel Point in Bowstring Truss 

of Fig. 54. 

One-third of the dead 
panel load, or 5 730 pounds, is taken as acting at the upper 
panel points, while the remainder, 11 470 pounds, acts at the lower 


Fig: 61. Fig. 62. 

Analysis of Stresses in Various Members of the Bowstring Truss of Fig. 54. 



Analysis of Stresses in Various Members of the Bowstring Truss of Fig. 54. 


ones. Articles 27, 28, and 29 should be carefully reviewed before 
going further. The shear times the secant method cannot be con- 
veniently employed for the live-load stresses in the members U t L 2 
and LJJ 2 , as the section will cut the member UJJ 2 , and the vertical 
component of its stress must be reckoned with in the stress equation 
The method of moments as illustrated in Fig. 27, Article 29, will be 
used for these members. 

The dead-load reaction is 2 X 17.2 - +34.4. The dead-load 
chord stresses should first be computed. 

By resolving the horizontal forces around L v it is seen that L L l 
= LjL 2 (see Fig. 55). Passing the section a a, taking the center 
of moments, at U v and stating the equation of the moments to the 
left of the section, there results (see Fig. 56) : 

+ 34.4 X 20 - L,L 2 X 20 = .-. L,L 2 = +34.4 

For LJjy the section b b is passed; the center of moments is 
at U 2 ; and the equation of the moments to the left of this section 
(see Fig. 57) is: 
+ 34.4 X 2 X 20 - (11.47 + 5.73) 20 - L 2 L 3 X 25 = .'. L 2 L 3 = +41.26. 

By passing a vertical section cutting L^ and LJJ V the stress in 
L U l can be determined by taking the sum of the vertical forces to 
the left and equating them to the vertical component of the stress 
(see Fig. 58). The equation is: 
+ 34.4 + L U 1 X 0.707 = 0, from which L U 1 = -34.4 X 1.414 = - 48.7. 

A section a a (Fig. 59) shows that the center of moments for 
UJJ 2 is at U 2 ; and stating the moments of the stress, and the forces 
to the left of the section, there results an equation in which an 
unknown lever arm enters. This lever arm I is readily computed 
to be 24.28 feet, and the equation can now be written: 

+ 34.4 X 2 X 20 - (11.47 + 5.73) 20 + UJJ* X 24.28 = 
.'. U,U 2 = -42.51. 

The stress in U 2 U 3 is determined by passing a vertical section 
in the 3d panel, and taking the sum of the horizontal forces. As there 
is no dead-load stress in the members L 2 U 3 and UJL Z , their compo- 
nents will be zero. Therefore (see Fig. 60) it is evident that U 2 U 3 
must be equal and opposite to L 2 L 3 and will be equal to 41.26. 

By reference to Fig. 55, the stress in L 1 U l is seen to be tensile and 
equal to +11.47. 


Pass a circular section around U 2 and take the sum of the vertical 
components, assuming that the stress in UJL 2 acts away from the 
section. The length of UJJ 2 is 1/5 2 + 20 2 = 20.6, and therefore the 
vertical component of UJJ 2 will be (42.51 H- 20.6) X 5 = 10.32, 
which acts upward. The stress equation of U^L^ (see Fig. 61) is: 

+ 10.32 - 5.73 - U 2 L 2 = .'. U 2 L, = +4.59, 

showing that a tensile stress occurs in U 2 L 2 when all panel points are 

The simplest method of ascertaining the stress in U^ is to pass 
a vertical section cutting members as shown in Fig. 62, and to equate 
the horizontal forces and stresses. The horizontal component of 

~ X 20 = 41.30, which acts toward the left. 

The equation of stress is, then : 

-41.30 + 34.40 + C7,L 2 sin = 0; but sin < = 0.707; 
t/,L 2 = +6.90 X 1.414 
= +9.76 

All the dead-load stresses being computed, the next operation 
will be to determine the live-load chord stresses. These are pro- 
portional to the dead-load stresses in the same ratio as the live panel 
load is to the dead panel load. This ratio is 50 -f- 17.2 - 2.907, 
and the chord and end-post live-load stresses are: 

Lot/, = -48.71 X 2.907 = -141.7 

U 1 U 2 = -42.51 X 2.907 = -123.6 

U 2 U 3 = -42.26 X 2.907 = -123.0 

L L 2 = +34.40 X 2.907 = +100.2 

L 2 L 3 = +41.26 X 2.907 = +120.3 

Also, the stress in U 2 L 2 when the live load covers the entire bridge 
is not 2.907 X 4.59, as it must be remembered that part of the dead 
load is at the panel points of the upper chord. Taking a circular 
section around U 2 (see Fig. 61), and noting that there is no load at U 2 , 
it is seen that the stress in U 2 L 2 due to live load is simply equal to the 
vertical component of the live-load stress of UJ7 2 and wjll be tensile. 
It is: 

U 1 U 2 = (123.6" -J- 20.6) X 5 = +30.0. 

The maximum live-load stress in U l L l is tensile, and equal to 
the live panel load at L l (see Fig. 55). 



To obtain the maximum stress in U 2 L 3 , load L 3 and L 4 . The 


shear F 3 will then be (1 + 2) - +30.0. The section will cut 

the members as shown in Fig. 63, and the equation of stress will be : 

+ 30.0- UtLa cos <=0; but cos = -z==r =0.782; 

.'. C/^ = +38.4. 

If panel points L l and L 2 were loaded, it is evident that the stress 
in L 2 U 3 would be +38.4. 

To obtain the maximum live-load stress in VJj v a section is 
passed cutting UJ7 V L^L V and U,L 2 (Fig. 64). The center of 
moments will be at the intersection of UJJ 2 and L,I/ 2 , and this point 
lies some place to the left of the support L . The lever arm of U t L 2 
will be the perpendicular distance from this point to the line U t L 2 
extended. The panel points L 2 , L 3 , and L t are loaded. The left 


reaction is then (1 + 2 + 3) = + 60.0. The lever arms are 


easily computed, and these, together with the members cut, are shown 
in Fig. 64. The equation of stress is : 

- 60.0 X 60.0+ t7 1 L 2 X>70.8 = .-. U,L 2 = +50.80. 

If a load were put on L l only, then the reaction at L would be 

X 50 = 40; and the equation of stress would then be: 

-40.0 X 60.0 + 50 X (60.0 + 20.0) + U,L 2 X 70.8 = .-. C/,L 2 = -22.6. 

As this is compression and greater than the dead-load stress, + 9.76, 
a counter is required in that panel. In order to get the stress in the 
counter, it must be inserted, U^ being left *out, and the dead and 
live load stresses computed and their difference taken. Fig. 65 gives 
the lever arms, center of moments, and the forces acting in this case. 
The dead-load stress is: 

-34.4 X 60.0 + (11.47 + 5.73) (60 + 20) -, X 62.5 = 
.'.L^U, = -11.02; 

and the live-load stress is: 

-40 X 60 + 50 (60 + 20) - LJJ^ X 62.5 = .-. L 1 f/ 2 = +25.60, 

and the stress in the counter is the algebraic sum of these two, or 
-11.02 + 25.60 - +14.58. 



When a live panel load is at L v LJJ 2 is acting, as has just been 
proved. As this load at L l causes a negative shear in all panels to 
the right, this negative shear in the center panel will cause L 2 U 3 to 
act. A section may now be passed as shown in Fig. 66, and the stress 
equations for U 2 L 2 written: 

For dead load, +34.4 - 11.47 - 2 X 5.73 - U 2 L 2 = .'. U 2 L 2 =- .-11.47 
For live load, +40.0 - 50.0 - U 2 L 2 = .'. U 2 L 2 = -10.00 

Total = + 1.47 

This is evidently not a maximum for U 2 L 2 , for when a full live load 
was on the span, the stress was +30.0 due to live load and +4.59 
due to dead load. 

It might be well to consider what effect is produced by loading 
L 3 and L 4 . The loading of L 2 and L l need not be considered, since 
it is evident that, as this causes the total shear in panel 2 to be positive 
and the total shear in panel 3 to be negative, therefore UJj 2 arid L 2 U 3 

Pig. 68. Stress Diagram of Half-Span of Parabolic Truss of Fig. 54. 

will act, and this causes a tensile stress in U 2 L 2 equal to the vertical 
components of the dead and live load stresses in UJJ 2 less the dead 
panel load at U y With a live panel load at L 3 and L v the left reaction 

is __(! + 2) = +30.0. The section, the live-load forces, the cen- 
ter of moments, and the members acting are shown in Fig. 67. The 
dead -load stress in U 2 L 2 will be the same as when the truss has no 
live load on it. The stress equation for the live load is: 

-GO X 30 - (60 + 20 + 20) X L 2 U t = .'. L 2 U 2 = -18.0. 




The dead-load stress being +4.59, this 
stress of 18.0 causes a reversal of stress 
in the vertical. For this reason the ver- 
ticals of bowstring trusses are, like web 
members of Warren trusses, built so as 
to take either tension or compression. 
The minimum stresses in the diagonals 
will be zero, for when one diagonal in a 
panel is acting, the other is not. 

The diagram of half of the truss in Fig. 
68 gives all the stresses. 

It is to be noted by the student, that 
in some cases one method for the deter- 
mination of stresses is preferable to others 
in that it saves labor of computation. 
The analysis of the truss of Fig. 68 illus- 
trates this fact. 

42. The Baltimore Truss. Baltimore 
trusses are of two classes those in which 
the half-diagonals, called sub-diagonals, 
are in compression, and those in which 
the sub-diagonals are in tension. The 
latter class is the one most usually built, 
as it is more economical on account of 
many of its members being in tension, in 
which case these members are cheaper 
and easier to build than if they were com- 
pression members. Fig. 14 shows botii 
types of truss. The Baltimore truss does 
not have a simple system of webbing, and 
t'oi that reason the analysis is here pre- 
sented. As the tension sub-diagonal truss 
is the type in most common use, its analy- 
sis will be given. 

Let it be required to compute the 
maximum stresses in the 14-panel 280- 
foot span of Fig. 69. The height is 40 
feet, the dead panel load 24 000 pounds, 



and the live panel load 40 000 pounds. One-third of the dead 
panel load is applied at the upper ends of the long verticals and also 
of the half-verticals. These half-verticals are designated as sub- 
verticals. Attention is called to the system of notation used for the 
ends of the sub-verticals. The full lines in Fig. 69 represent the 
main members, being stressed by dead load only. The heavy lines 
indicate those members that take compression, the light lines those 
that take tension, and the broken lines the counter-braces. In this, 
as in nearly all Baltimore trusses, the diagonals make an angle of 45 
degrees with the vertical. 

The dead and the positive live-load shears in the various panels 
should be computed. They are : 


V l +156.00 P! = (1 + 13) Z_ = +260.00 

V 2 +132.00 Y 2 = (1 + ....12)~ = +223.00 

Y 3 + 108.00 V 3 = (1 + .... 11) ^ = + 188.50 

F 4 + 84.00 F 4 = (1 + 10)-^ = +157.20 

F 5 + 60.00 V 5 = (1 + .... 9) ~ = +128.50 

F 6 + 36 00 F 6 = (1 + 8) y = + 102.80 

V 7 + 12.00 V r = (1 + .... 7) |5. = + 80.00 

It is only necessary to determine the negative live-load shear in 
panels 5 and 7, in order to ascertain if there is a counter required. 
These shears are : 

- V s = (10 + 11 + 12+ 13) - 4 X 40 = -28.60 

-F 7 = (8 + 9 + 10 + 11 + 12 + 13) y^- - 6 X 40 = -60.00 

From a comparison of these with the dead-load shears, it is seen 
(see Article 37) that a counter is required in panel 7 only. 

The dead-load stresses are first to be computed. The stress in 
any sub-vertical is found by passing a circular section around its 
lower end, and equating the sum of the vertical forces, assuming 
in this, as in all cases, that the unknown stress acts away from the 




section. Take 3/jTO,, for example. Fig. 70 gives the section, the 
forces acting, and the members cut. Then, 

+ M l m 1 - 16.0 = .'. M imi = +16.0 

As all sub-verticals have the same dead load at their lower end, it 
follows that the dead-load stress in all sub-verticals is the same, a 
tensile stress of 16 000 pounds. 

The dead-load stresses in the sub-diagonals are determined by 
resolving the forces around the joint at their lower end. The com- 
ponents perpendicular to the diagonal are taken (see Fig. 71). Take 

Fig. 70. Diagram for Calculating Stress in 
Sub- Vertical of Baltimore Truss. 

m 2 f7 2 . The known forces or 
stresses are the dead panel load 
of 8.0 and the stress in m 2 M 2 , 
which is 16.0 and which being 
tensile acts away from the sec- 
tion. The stress equation is: 

Fig. 71. Diagram for Calculating Stress in 
Sub-Diagonal of Baltimore Truss. 

+ m 2 U 2 - 8.0 sine <f> - 16.0 sine = 0. 
c = 45, sine = 0.707,and 
m 2 U 2 - 8.0 X 0.707 - 16.0 X 0.707 = 
.\m,U a = +16.96. 

This equation may be put in another form by multiplying and dividing 
the numerical values by 2: 

(8.0 + 16.0) 

X 1.414 = 0; 



which proves the well-known saying that the stress in the sub-diagonals 
w equal to one-half the panel load, times the secant of the angle <f>. It 
also shows that the vertical component of the sub-diagonal is equal to 



one-half 'the panel load. This fact should be remembered, as it will 
be frequently used further on. 

In a similar manner, the stress in all the tension sub-diagonals 
will be found to be the same, + 16.96, and 
the stress in the compression sub-diagonal 
mjji is -16.96. 

The stress in the member L m l and in 
the upper half of any main diagonal (i. e., 
Ujrn v Ujn 3 , and ?7 3 m 4 ) is determined as in 
the diagonals of the Pratt or Howe truss, 
for the section passed cuts but one mem- 
ber, which has a vertical component. Take 
L m l (see Fig. 72). Then +156.0 + L Q m l 
cos 45 = 0, from which L m l = 220.5. For U^ the section is 
passed as in Fig. 73, and the equation of stress is + V 3 Ujm 2 cos 
45 = 0, or +108.0 - U^ X 0.707 = 0, from which Z7,ra 2 = 
+ 152.9. 

In a similar manner, 

U 2 m 3 = +60.0 - 0.707 = +84.84; 
U 3 m 4 = +12.0 + 0.707 = +16.96. 

The stresses in m l U l , w 2 L 2 , and m 3 L 3 may be determined by 

Fig. 72. Stress in Diagonal 
of Baltimore Truss. 

Fig. 73. Stress in Upper Half of Main 
Diagonal of Baltimore Truss. 

Pig. 74. Stress in Diagonal of Baltimore 

resolving the forces about m v m v and ra 3 ; but a neater solution is to 
pass a vertical section cutting the member whose stress is desired, 
and to equate to zero the shear and the vertical components of all 




the members cut (see Fig. 24, Article 28). The section for m 1 U i is 
passed as in Fig. 74. The equation of stress is then : 
m l U l cos 45 + m,!/! cos 45 + F 2 = 0; 

but the vertical component of m^ is = 12; and therefore, 

m l U l X 0.707 + 12 + 132 = 
.'. m l U l = -203.6. 

For m^L 2 , the section is as shown in Fig. 75, and the stress 

equation is : 

-m 2 L 2 X 0.707 + vert, component m 2 U 2 + V t = 
-m^ X 0.707 + 12 + 84 = 
.-. m 2 L 2 = + 135.6. 


[Fig. 75. Calculating Stress in Lower Half-Diagonal of Baltimore Truss. 

In a similar manner, passing a section cutting U 2 U 3 , m 3 U 3 , mJL^ 
and M^Ly the stress equation may be written: 
- m 3 L 3 X 0.707 + 12 + 36 = 
.-. m 3 L 3 = +67.85. 

The stresses in the verticals are best determined by resolving 
the vertical forces at their lower end. Referring successively to 
diagrams a, b, and c of Fig. 76, the stress equations are : 

+ U 1 L 1 - 16.0 - 12.0 = .'. C7,L, = +28.0 

+ U 2 L 2 - 16.0 + 96.0 = .'. U 2 L 2 = -80.0 

+ U 3 L 3 - 16.0 + 48.0 = .'. U 3 L 3 = -32.0 

96 and 48 being the vertical components of ra^ and m 3 L 3 respec- 




The chord stresses are easiest computed by considering the 
resolution of horizontal forces at the panel points. As the diagonals 
make an angle of 45 with the vertical, their horizontal and vertical 

16.0 16.0 

Fig. 76. Calculating Stresses in Verticals of Baltimore Truss of Fig. 69. 

components are equal. For instance, the horizontal component of 
the members Ljn v V \m v and U 2 m 3 are equal to the shear in that 
panel, which is their vertical component. At point I/ (see Fig. 77), 
there results: 

+ L Afj horizontal component of AfjL = 0; or, 

+ L M l 156 = 

/. L M, = +156.0; 

and from Fig. 70 it is evident that L M t = 
Fig. 78), L t M 2 is equal to M l L l , 
less the horizontal component of 
MJj v and the equation is : 

-156 + 12 + L,M 2 = 
. L,M 2 = + 144.0; and M 2 L 2 = + 144.0. 

At point L 2 (see Fig. 79), L 2 M 3 
is equal to the sum of the horizon- 
tal components of MJL 2 and ra^; 
that is, 

+ L a Af, - 144.0 - 96.0 = 

.-.L 2 M 3 = M 3 L 3 

At point Z/j (see 


Fig. 77. Chord Stress in Baltimore Truss. 

+ 240.0. 
In a similar manner, at point L 3 , the stress equation is: 

+ L,M 4 - 240.0 - 48.0 = 

.-. L 3 M 4 = M 4 L t = +288.0. 

At the upper panel point U l (see Fig. 80), there results the 

equation : % 

+ U l U t + hor. comp. f7jOT 2 + hor. comp. m,?7, = 0; 
17,17, + 108.0 + 144 = 0; or, ?7,t7 2 = -252.0. 
For the member U 2 U 3 (see Fig. 81), the equation is: 




+ Ujj^ 4. U 2 U 3 hor. comp. m 2 U 2 + hor. comp. U 2 m 3 = 
+ 252.0 + U 2 U 3 - 12 + 60 = 
.-. U 2 U 3 = -300.0. 

In a similar manner, by resolving the horizontal forces at U 3 , it 
will be seen that the action of m 3 U 3 will neutralize that of U z m v as 

L , m *^ L 2 

Fig. 78. Fig. 79. 

Bottom Chord Stresses in Baltimore Truss. 

they are equal and pull in opposite directions, and U a U t is equal to 
U 2 U 3 = -300.0. 

The live-load stresses in the chords, the end-post, and the sub- 
diagonals are all proportional to the dead-load stresses in the same 

Fig. 80. Fig. 81. 

Top Chord Stresses in Baltimore Truss. 

ratio as the live panel load is to the dead panel load. This ratio is 


= 1.667. By reference to Fig. 70, it will be seen that the live- 
load stress in the sub-verticals is +40.0 for each one. The following 
stresses can now be determined : 

L omi = -220.5 

m 1 U l = -203.6 

U,U 2 = -252.0 

U 3 U a = -300.0 

U 3 L\ = -300.0 

L L, = +156.0 

L,L 2 = +144.0 

L 2 L 3 = +240.0 

L 3 L 4 = +288.0 

X 1.667 = -367.5 

X 1.667 = -339.5 

X 1.667 = -420.0 

X 1.667 = -500.0 

X 1.667 = -500.0 

X 1.667 = +260.0 

X 1.667 = +240.0 

X 1.667 = +405.0 

X 1.667 - +481.0 



m.L, = - 16.96 X 1.667 = - 28.28 
m a U 3 = m 3 U 3 = +16.96 X 1.667 = +28.28 

The vertical U^ will have its maximum live-load stress when 
points M 1 and L l are loaded, for these are the only loads which cause 
a stress in that member (see Fig. 76a). The equation is: 

~ - 40 + 

= 0, 

from which, 
C/,L, = +60.0. 

The maximum 
live-load stresses 
in Ujn v U 2 m 3 , and 
U 3 m 4 are obtained 
in a manner exact- 
ly like that used 
in obtaining dead- 
load stress, only 
the live-load posi- 
tive shear is used. 
The stresses are: 


Stress in Lower Half of Main Diagonal of Baltimore 

U,m a X 0.707 -f- 188.5 = 
U 3 m~ 3 X 0.707 + 128.5 = 
U 3 m t X 0.707 + 80.0 = 

.'. U l m 2 = +2665 
. . U 2 m 3 = +181 5 
.'. U 3 m t = +113.1 

In the determination of the maximum live-load stress in the 
lower halves of the main diagonals, m 2 L 2 , ra^, and m 4 L 4 , one of the 
peculiarities of this truss becomes apparent. A section being passed 
as in Fig. 82, the panel point ahead of the section, and all between 
the section and the right support, must be loaded. This of course 
produces a stress in m 2 U 2 , and the vertical component of this enters 
the stress equation." The shear in the section a a under this load- 
ing is: 

V._. = + 188.5 - 40 = + 148.5; 

and the stress equation is: 

-ijL a X 0.707 + -^ + 148.5 = 
.-. m 2 L 2 = +238.0. 

If the truss had been loaded from the section to the right, there being 
no load on M 2 , no stress would result in m 2 U 2 , and the stress in ra 2 L 2 




157 2 
would have been mJL 2 = ' = +222.2. In a similar manner, 

by loading successively points M 3 and to the right, and M t and to 
the right, the stress equations of m^ and ra 4 L 4 are : 

-m 3 L a X 0.707 +^- + 128.5 - 40 = .'. m 3 L 3 = +153.3 

-m 4 L 4 X 0.707 + -~- + 80.0 - 40 = .'. m 4 L 4 = + 84.8 

The maximum live-load stresses in the main verticals occur when 
the panel points to the right of the section which cuts the member 

under considera- 

II. 0\ ^ 1 I - 

tion are loaded. 
There being no 
load at the end of 
the sub-vertical 
just to the left of 
the section, there 
will be no stress 



Fig. 83. Stress in Main Vertical of Baltimore Truss. 

in the sub-diag- 
onal which the sec- 
tion cuts. The chords, of course, do not exert a vertical com- 
ponent; and so the only unknown term of the stress equation is the 
stress in the member itself. Fig. 83 show r s how the section should 
be passed when U^L 2 is considered. The stress equation is: 
+ U 2 L 2 + F a _ a = 0; + U 2 L 2 + 128.5 = 0; .-. U 2 L 2 = -128.5. 

In a similar manner, by passing a section cutting U 2 U 3 , m 3 U 3 , 
UgLg, L 3 m t , and loading M t and to the right, it is seen that the stress 
equation for UJj 3 is: 

+ U 3 L 3 + 80.0 = .'. U 3 L 3 = -80.0 

The components of m 3 U 3 and L 3 m 4 are zero, as can readily be proved 
by solving for them under this loading. 

Fig. 84 gives all the stresses, and they are written in order of 
dead load, live load, and maximum. 

43. Other Trusses. The analysis of the foregoing trusses will 
enable one to solve any of the trusses of modern times. For the 
solution of the Whipple (sometimes called the "double-intersection 
Pratt") and others which are not mentioned in this text, the student 



is referred to the text- 
books of F. E. Tur- 
neaure and Mans- 
field Merriman. 


44. Use of En= 
gine Loads. It was 

formerly the custom 
for railroads to spec- 
ify that the engine to 
be used in computing 
the stresses in their 
bridges should be one 
of their own which 
was in actual use. 
The engines of differ- 
ent roads were usual- 
ly different both in 
regard to the weight 
on the various wheels 
and in regard to the 
number and spacing 
of the wheels. Of 
late years, consider- 
able progress has 
been made towards 
the adoption of a 
typical engine load- 
ing as standard. 
These typical engines 
(see Fig. 17, Article 
25) vary only in re- 
gard to the weights 
on the wheels, the 
number and spacing 
of wheels being the 
same in all engines. 



- c r; ;= 


The distance between wheels is an 
even number of feet, instead of an odd 
number of feet and inches and frac- 
tions thereof. For examples of load- 
ings which are in almost universal use, 
consult the specifications of Cooper or 

The labor of computation of stresses 
when engine loads are used is consid- 
erably lessened by the use of the so- 
called engine diagrams. Fig. 85 gives 
a diagram which has been found very 
convenient. The first line at the top 
represents the bending moment of all 
the loads about the point to the right 
of it. All the loads are given in thou- 
sands of pounds, and all the moments 
are in thousands of pound-feet. The 
practice of writing results in thousands 
of pounds or, in case of moments, in 
thousands of pound-feet or pound- 
inches is to be recommended, as it 
saves the unnecessary labor of writing 
ciphers. Throughout this text this 
practice has been extensively followed, 
the stresses being written to the near- 
est ten pounds or one-hundred pounds 
as the case may be. For example, 
6 433 may be written 6.43 or 6.4, the 
few pounds which are neglected mak- 
ing no appreciable difference in the 
design. The distances are in feet. 

As an example of the use of the first 
line at the top, suppose that it is de- 
sired to find the moment of all the 
loads to the left of a certain point 
when wheel 6 (the numbers of the 
wheels are placed inside of the circles 



representing the wheels) is just over the point. The moment will 
be 1 640 000 pound-feet, which is obtained by reading off the 1 640 
just to the right of the line through wheel 6. 

When using the first line for values at sections in the uniform 
load, the values given represent the moment of all wheel and uniform 
loads about the points in the line or section to the left of the value 
given. For example, if it is desired to find the moment about a 
point in line 2, it will be 19 304 000 pound-feet, the value 19 304 
appearing to the right of the line. 

The line of figures below the wheels indicates the distances 
between any two wheels. 

The third line of figures indicates the distance from the first 
wheel to the wheel to the right. For instance, 37 is the distance from 
wheel 1 to wheel 7, 

The values in the fourth line indicate the sum total of all the 
loads to the left of the value given. For example, 245 signifies that 
the loads 1 to 15 inclusive weigh 245 000 pounds. 

The values in lines 5 and 6 are similar to those of lines 3 and 4, 
except that the starting point is at the head of the uniform load. 
For example, 40 in line 5, and 112 in line 6, indicate that it is 40 feet 
from the head of the uniform load to the wheel 12, and that wheels 
18 to 13 inclusive weigh 112 000 pounds. 

The values in lines 7 to 16 indicate the value of the moment of 
all the wheels from the zigzag line up to and including the one to the 
left or the right, according as the value is to the left or the right of 
the zigzag line. For example, 2 745, line 11, indicates that the 
moments of wheels 8 to 14 inclusive about the zigzag line just under 
wheel 15, is 2 745 000 pound-feet; or the value 1 704, line 14, shows 
that the moments of wheels 13 to 18 about the zigzag line just under 
wheel 12 is 1 704 000 pound-feet. 

When line 4 of figures is under the uniform load, the values refer 
to the vertical line to the right; 'thus 324 is the value of all loads to 
the left of line 3 about that line. 

For values of moments at points which fall in between wheels, 
or at positions in the uniform load where the value of the moment 
is not given, a very important principle of applied mechanics is used. 
It is: 

M. = M' + Wx +. '. 





M a = Moment at section desired; 
M' = Value of moment at preceding vertical line; 
W = Sum total of all loads to the left of and at the point where M' 

is taken; 
x = Distance from section under consideration to vertical line to 

which M' is referred; 
w = Uniform load on the distance x. 

Let it be desired, for example, to determine the moment at a 
a point c, 3 feet to the right of wheel 

13. The position of the loads is 
given in Fig. 86. The moment is: 

Ma-. = 7 668 + 212 X 3 
= 7 668 + 636 

= 8 304 = 8 304 000 pound- 
feet, there being no uni- 
form load. 

To illustrate the method when 
applied to points in the uniform 
load, assume the point to be 7 feet 
to the right of line 2. The po- 
sition is illustrated in Fig. 87. The 
moment is : 














Fig. 86. Calculation of Moment at a Point 
under Engine Load. 

= 19 304 + 304 X 7 + 

7 2 X 2 

= 21 481 = 21 481 000 pound-feet. 

The use of the moment diagram is now apparent. Reactions 
due to any position of the engines may be determined by dividing the 
span into the value obtained for the moment at the right end of the 
span. Likewise, if the moment of the reaction about any panel point 
is determined and from it the moment of the wheel loads about that 
same panel point are subtracted, then the result, divided by the 
height of the truss, will give the chord stress. For example, if the 
right end of an 8-panel 196-foot span truss, height 25 feet, came 7 
feet to the right of the vertical line 2, then the moment at this point 
(see Fig. 87) would be 21481000, and the reaction would be 
21 481 000 -T- 196 = 109 600. This position of the loads would 
cause the panel point L 6 to come 3 feet to the right of wheel 13. The 
moment of the reaction about L R is 109 600 X 6 X 24.5 = 16 111 200; 



and the chord stress U 6 L 6 for this position of the engine is: 

16111200 - 8304000 

- 312 000 pounds. 

In using the engine to determine the shear in any particular 
panel, it must be remembered that the shear is not the left reaction 
less all the loads to the left of the panel point on the right of the 
section, as the loads in the panel under consideration are carried on 
stringers, and these stringers transfer a portion of the loads to the 


ie3 64 





per L incorFootx/x 







Fig. 87. Calculation of Moment at Point under Uniform Load. 

panel point on the left of the panel, and a portion to the panel point 
on the right of the panel. Only that portion of the loads in the panel 
which is transferred to the left panel point should be subtracted from 
the reaction, as should all of the loads to the left of the panel under 
consideration. If, in a 6-panel 120-foot span Pratt truss, the wheel 
6 comes at L y , the left reaction will be : 


16 364 + 3 X 284 + 

3 2 X 2.0 


and the loads in the first two panels will be in position as indicated 
by Fig. 88, the wheel 3 being 1 foot to the right of point L r Let it 
be required to determine the shear in the panel L^ 2 when the loads 
are in this position. It will be the reaction 143.6 minus loads 1 and 
2 and also that portion of the loads 3, 4, and 5 which will be trans- 
ferred by the stringers to -the point L r As the stringers are simple 








Fig. 89. Shear Diagram for Panel under 
Engine Load. 

beams, the amount transferred to L t will 
be the reaction of the stringer L^L r Re- 
ferring to Fig. 89, the reaction is: 
R Li = (20 X 9 + 20 X 14 + 20 X 19) + 20 
= 42.0 

The shear in the second panel is now 
found to be : 

V 2 = 143.6 -(10 + 20 + 42.0) = +71.6 
In the majority of cases where it is 
necessary to determine the shear in a 
panel, none of the loads will be in the 
panel to fhe left of the one under consid- 
eration. In this case the operation is 
somewhat simplified, as the engine dia- 
gram can be used directly. If the engine 
be placed so that the third wheel is at 
L v wheel 16 will be just over the right 
support, and the left reaction will be : 
Ri = 12 041 * 120 = 100.3. 

As there are no wheel loads in the first panel, the amount to be sub- 
tracted from the reaction will be that proportion of the loads 1 and 2 
which is transferred to I,,; and this (see Fig. 90) is 230 -=- 20 = 11.5. 
The shear in the second panel is then 100.3 - 11.5 = +88.8. 

From inspection of the resulting shear in the second panel when 
wheel 6 is at L 2 and when wheel 3 is at L 2 , it is seen that different 
wheels at L 2 will give different shears in the panel to the left. Evi- 
dently there is some wheel which will give the greatest shear possible. 
The same is true of the relation between wheels and moments- The 


next two articles are devoted to subject-matter which will enable. one 
to tell which of several wheels is the correct wheel at the point, without 
the necessity of solving for the shear each time every wheel is at the 

45. Position of Wheel Loads for Maximum Shear. By methods 
of differential calculus, it can be proved that, for any system, either 
of wheel loads or wheel loads followed by a uniform load (see Fig. 91), 
the correct wheel that should be at the panel point b in order to 

6 at eo'=ieo' 

Fig. 90. Determination of Shear in Panel under Engine Load. 

give a very great or maximum shear in the panel a b, is such a 

W W 

wheel that the quantity Q = - G is positive when q = - 

m m 

(G + P) is negative. In these equations, 

W = Total load on the truss; 

m = Number of panels in the truss; 

G = Load in panel under consideration; and 

P = Load at panel points on right of panel. 

If a load is directly over the panel point a, it is not to be included in 
the weight G; neither is P included in the weight G, If a wheel load 
should come directly over the right end of the truss, it should not be 
considered in the quantity W. 

The only way to determine which wheel is the correct one, is 
to try wheel 1, then wheel 2, and so on, until the wheel or wheels are 
reached that will give the Q and q signs of an opposite character. 




The process should not be stopped there, but the next succeeding 
wheels should be tried until Q and q again have the same sign. 

As an example, let it be required to determine the position of the 
wheel loads to produce the maximum positive shears in a 6-panel 
120-foot Pratt truss. This work should be arranged in tabular form, 
and Table V is found to be convenient. 


Determination of Position of Wheel Loads for Maximi 










G + P 






if - 47.33 





L t 



if = 48.67 





gives a maximum 




if _ 50.03 





gives a maximum 




if - 50.03 





L 2 



~j^- = 38.67 





L 2 



-^ = 40.83 





gives a maximum 

L 2 



^p- = 43.00 





L 3 


-IT - 25 ' 67 





L 3 



-^p = 28.67 





gives a maximum 


L 3 



T - 34 ' 





gives a maximum 


L 3 



TT- = 37 - 





L 4 






L 4 



-i^- = 21.50 





gives a maximum 




-l^ 2 . = 23.67 







- - = 11.67 





L & 



= 15.00 





gives a maximum 




-^L - 17.17 








</) PS 


5 5 

2 i 

H . 


X > 
Jfl * 



A study of Table V shows the fact 
that wheel 1 can never produce a maxi- 
mum. It also shows that there are in 
some cases two positions which will give 
large values of the moment. In these 
cases the shears for each position of the 
engines must be determined in order to 
tell which wheel at the panel point in 
reality gives the greatest. In practical 
work it is customary to use the first 
position found, as the difference in the 
shears resulting from the use of the two 
positions is not large enough to affect the 
final design. 

Fig. 92 shows the engine diagram 
on the truss in the correct position to 
give the maximum shear in the second 
panel. The weight of wheel 16 is not 
included in the weight W, as it is directly 
over the right support. 

46. Position of Wheel Loads for 
Maximum Moments. In this case the 
methods of differential calculus are em- 
ployed to determine which wheels will, if 
placed at a point, give a maximum mo- 
ment at that point. For any system, 
either of wheel loads or of wheel loads 
followed by a uniform load, that wheel 

which will cause K L to be positive, and k = -^-^ 

m m 

(L + P) to be negative, is the wheel. Here n is the number of the 
panel under consideration, and is to be reckoned from the left 
end; L = the load to the left of the point under consideration; 
and the remainder of the letters signify the same as they do in 
Article 45. In some cases there will be more than one position 
of the loads which will satisfy the above condition. It is then 
necessary to work out the actual moments created by the loads 
in each position, in order to find out which is the largest. The 




position of the loads 
for the greatest mo- 
ments should be de- 
termined for all panel 
points except the one 
on the extreme right, 
as the greatest moment 
possible maybe caused 
by wheels of the rear 
engine being on the 
point on the right- 
hand side of the truss, 
instead of the wheels 
of the front engine be- 
ing at the correspond- 
ing point on the left- 
hand side. 

In general, it may 
be said that there will 
be a number of wheels 
which, if placed at the 
panel point in the cen- 
ter of the span, will 
satisfy the given con- 
ditions. In this par- 
ticular case, it is not 
necessary to determine 
all of the moments. 
The greatest moment 
possible will occur 
when that one of the 
heaviest wheels of the 
second locomotive 
which gives the heav- 
iest load upon the 
truss is at the point. 
In case several of the 
heavy wheels give the 



same maximum load W, use the first wheel which gives this max- 
imum W. 

Let it be required to determine the position of the wheel loads 
for maximum moments at the lower panel points of the 6-panel 120- 
foot Pratt truss of Article 45. The necessary work can be con- 
veniently arranged in the form of a table, as is done in Table VI. 

Determination of Position of Wheel Loads for Maximum Moments 









.+ ! 







284 -5- 6 = 47.3 
292 -5- 6 = 48.7 
302 -f- 6 = 50.3 
302 -5- 6 = 50.3 






gives a maximum 
gives a maximum 
wheel 1 is off bridge 






(271-5-6)X2 = 90.3 
(290--6)X2 = 96.7 
(300-r-6)X2 = 100.0 





gives a maximum 





(271n-6)X3 = 135.5 

(298 4-6) X3= 149.0 
(304 -5-6) X3= 151.3 
(294n-6)X3 = 147.0 
(274-h6)X3 = 137.0 






gives a maximum 
gives a maximum 
gives a maximum 
gives a maximum 
gives a maximum 

L 4 






(271 -=-6) X4 = 180.6 

(294-=-6)X4 = 196.0 
(304-r-6)X4 = 202.6 






Note wheel 18 not 
gives a maximum 
gives a maximum 

One should carefully note that in certain positions, as when 
wheels 11, 12, 13, and 14 are at L 3 , some wheels are to the left of the 
left support; that is, they are not upon the bridge. In all such cases 
they are counted neither in the quantity L nor in W. 

In the case of L 3 , wheel 11, being the first large driver of the 
second engine, will give the greatest moment, as it is the first driver 
to come at the point when the maximum load of 304 000 pounds is on 
the truss. Fig. 93 shows the engine diagram on the truss in correct 
position to give the maximum moment at point L r 

47. Pratt Truss under Engine Loads. In order to exemplify 
the use of the engine-load diagram, let it be required to determine the 
stresses in the Pratt truss of Article 45 due to E 40 loading, the 




height being 25 feet. The 
secant is (^? + ~20 2 )< -25 = 

The maximum positive 
shears in the various panels 
should first be computed. 
These, written in reverse or- 
der, will be the maximum 
negative or minimum shears. 
Table V should now be re- 
ferred to, and an outline dia- 
gram drawn to the same scale 
as the engine used, on which 
to place the engine diagram in 
the correct position. The vari- 
ous values can then be read 
off the diagram at the right- 
hand end of the truss. It will 
be found convenient to lay off 
to scale the first ten feet of 
the lower chord of the truss 
from the right support, mak- 
ing the divisions one foot apart. 
This will enable one to ascer- 
tain the distance of the last 
wheel load from the right sup- 
port, or the amount of uni- 
form load upon the bridge, 
without scaling or further com- 
putation. In case it is desired 
to have the wheel loads appear 
on the lower chord, as in Fig. 
93, the outline of the truss 
should be on tracing cloth or 
transparent paper. This is not 
to be advised, however, as er- 
rors are likely to occur because 
of failure to distinguish clearly 



the various numerical values. It is far better to place the diagram 
as in Fig. 92, in which case the outline of both the truss and the dia- 
gram can be drawn on good stiff paper. 

For wheel 3 at point L l (see Articles 44 and 45), the left reaction 
is as follows, there being four feet of uniform load upon the truss: 

(42 v 2 \ 
16 364 + 284 X 4 + * j -4- 120 = 146.0; 

and the proportion of loads in the panel which is transferred to 
the point L by the stringers is 230 -=- 20 = 11.5. The shear is 
therefore F t = +146.0 - 11.5 - +134.5. The computation for 
the shear when wheel 4 is at the point, will not be made; for, as has 
been noted before, the result will not be much different from the 

For wheel 3 at L 2 , wheel 16 comes over the right support. The 
left reaction is: 

Ri = 12041 -H 120 = 100.3; 

the proportional part of the loads which is transferred to L l is 11.5; 
and the shear is: 

V 2 = ( + 100.3 - 11.5) = +88.8. 

For wheel 2 at L 3 , wheel 11 comes four feet from the right sup- 
port. The left reaction is: 

Ri = (5 848 + 172 X 4) + 120 = 54.5. 

That part of wheel 1 which is transferred to L 2 is 80 -=- 20 = 4.0, 
and the shear is therefore: 

V 3 = ( + 54.5 - 4) = +50.5. 

For wheel 2 at L 4 , wheel 9 comes over the right support. The 
left reaction is: 

R\ = 3 496 + 120 = 29.1. 

That part of wheel 1 which is transferred to Z/ 3 is 4.0, and the shear 
is therefore: 

V 4 = ( + 29.1 - 4.0) == +25.1. 

For wheel 2 at L 5 , wheel 5 is five feet from the right support, 
and the left reaction is: 

Ri = (830 + 90 X 5) -T- 120 = 10.7 
and the shear is : 

F 5 = ( + 10.7 - 4.0) = +6.7. 



If the dead panel load is 20 000 pounds, all the shears may now 
be written as follows: 


+ 134.5 0.0 

+ 88.8 - 6.7 

+ 50.5 -25.1 

+ 25.1 -50.5 

+ 6.7 -88.8 

0.0 -134.5 

A comparison of the shears in the third and fourth panels shows 
that counters are required. The stress in these counters is: 
U 3 L 2 = U 3 L 4 = +1.28 X (25.1 - 10.0) = +19.2 

As it is known that positive shears cause a compressive stress 
in L U l and tensile stresses in the diagonals, and that negative 
shears produce compressive stresses in the intermediate posts, the 
left half of the bridge being considered, the web stresses for dead and 
live load can be determined without in all cases writing the stress 
equations in order to determine the sign. It should be remembered 
that one-third of the dead panel load, or 6 700 pounds, is applied at 
the panel points of the top chord. 

Dead-Load Stresses in the Diagonals 

L U t = -1.28 X 50 = -64.0 
7,1/2 = +1.28 X 30 = +38.4 
U 2 L 3 = +1.28 X 10 = +12.8 

Dead-Load Stresses in the Verticals. For Z7 2 Z/ 2 , the section 
passed will cut UJJ 2 , U^, and L 2 L 3 , and the shear on this section 
will be 50 - 2 X 13.3 - 6.7 = + 16.7. The stress equation is 
+ 16.7 + U 2 L Z = 0, from which UJ,, = - 16.7. 

The dead-load stress in U^ is found by passing a circular sec- 
tion around U y Then - U 3 L 3 - 6.7 = 0, from which U 3 L 3 = -6.7. 
In a similar manner, by passing a section around L v the stress is 
found to be + 13.3. 

Live-Load Stresses in the Diagonals 


LoU, = -1.28 X 134.5 = -172.2 

C7,L 2 = +1.28 X 88.8 = +113.6 -1.28 X 6.7 = -8.6 

U 2 L 3 = +1.28 X 50.5 = + 64.7 

Live-Load Stresses in the Verticals. The maximum stress in 
U l L l occurs when one of the large drivers is at L v and the loads ID 





o ( 

the first panel are as near as possible one-half the sum of the loads 
in panels 1 and 2 and the load at L r This can be established as a 
fact by use of the differential calculus. In the present case, this con- 
dition is satisfied when wheel 4 is at L r Then the weight of the 
wheels in panel 1 is 50 000 pounds, and the sum total is 116 000 pounds. 
If wheel 13 be placed at L v the result will be the same, and then the 
engine diagram can be used. Fig. 94 represents the engine diagram 
in place, ready to use. According 
to Article 44, the value 480 is the 
moment of wheels 10 to 12 about 
L v Therefore 480 -r- 20 (20 is 
the panel length) - 24.00, is that 
amount of wheels 10 to 12 which 
is transferred to L . In like 
manner, 529 -r- 20 - 26.45 is the 
amount of wheels 14 to 16 trans- 
ferred to L 2 . As the total weight 
of the loads in the two panels is 
116000 pounds, the amount 
transferred to L t must be 116.0 
- (24.00 + 26.45) = 65.55, and 
the stress in U^ is therefore 
+ 65.55. 

The maximum live-load stress 
in LJJ 2 occurs when the loading 
is in a position to give the maxi- 
mum shear in the third panel, as 
the shear at a section cutting UJJ V U 2 L 2 , and L 2 L 3 is the same' as 
that at a vertical section in the panel. The stress equation is + U 2 L 2 
+ 50.5 = 0, from which UJL^ = 50.5. In a similar manner, the 
stress equation for the maximum live-load stress in U 3 L 3 is + U 3 L 3 
+ 25.1 = 0, as U 3 L t is working, and therefore U^L 3 = 25.1. As in 
the case of the analysis of the Pratt truss under uniform load (see 
Article 39), the dead-load stress of 6.7 cannot be added to this 
stress of 25.1 to obtain the maximum; but the dead-load stress in 
U 3 L 3 must be obtained when diagonals U 3 L 3 and UJL,i are in action. 
In the manner explained in Article 39, this is found to be +3.30. 

It will be found that as the'engines come on the bridge from the 

Fig. 94. 

Engine Diagram for Determina- 
tion of Live-Load Stress in Vertical 

of Pratt Truss. 



left, the counters come into action in the case of U Z L 2 ; and in the 
case of U 3 L 3 , both U 2 L Z and LJJ^ act, thus causing the live-load stress 
in these verticals to be zero; and when this is the case, the dead-load 
stress is 6.7, which is the minimum. 

Dead-Load Chord Stresses. The dead-load chord stresses can 
be found by any of the methods previously given; but they will be 
found by the tangent method as indicated below, the tangent being 
20 -r- 25 = 0.8: 

LoL, = +0.8 X 50 = +40.0 = L,L 2 
UJJ, = -(50 + 30) X 0.8 = -64.0 
U 2 U 3 = -(50 + 30 + 10) X 0.8 = -72.0 
L 2 L 3 = -UiU, = -(-64.0) = +64.0 

Live-Load Chord Stresses. On account of the wheel loading, nc 
ratio can be established between these stresses and the dead-load 
chord stresses. The maximum moments at each point must be 
determined, and these divided by the height of the truss will give the 
chord stresses. For all points to the left of the center of the bridge, 
the main diagonal will act. For points to the right of the center, an 
uncertainty exists. The shear in the panels on either side of the 
point under consideration should be determined when the loading is 
in position to give the maximum moment at that point. This will 
indicate which diagonals act, which fact will indicate for what chord 
member that point is the center of moments. 

When wheel 3 is at L lt four feet of uniform load are on the truss, 
and the left reaction is: 

Ri = (16 364 + 384 X 4 + ^-~) + 120 = 146.0. 

The moment of this reaction about L v less the moment of wheels 1 
and 2 about L v will be the moment at L l due to this loading. The 
moment of wheels 1 and 2 about L l is taken from the diagram, where 
it occurs in the first line of values just to the right of the vertical line 
through wheel 3, and therefore: 

M l = 146.0 X 20 - 230 = 2 690 000 pound-feet. 

When wheel 4 is at L v there are nine feet of uniform load on 
the truss, and the left reaction is: 

Ri = (16 364 + 284 X 9 + ^~ ) -=- 120 = 158.0; 
and in this case, 

M l = 158.0 X 20 - 480 = 2 680 000 pound-feet. 


As this is less than when wheel 3 is at the point, wheel 3 gives the 
greatest moment. 

When considering the point L 2 with wheel 6, the left reaction is: 

02 y o 

Ri = (16 364 + 284 X 3 + -=) H- 120 = 143.5 
M 2 = 143.5 X 2 X 20 - 1 640 
= 4 100 000 pound-feet. 

The conditions at L 3 indicate that there are several wheels which 
give large moments; but according to Article 46, wheel 11 gives the 
maximum moment. When this wheel is at L 3 , wheel 1 is off the 
truss, and 15 feet of uniform load are on the truss. The moment of 

att the loads about the right support is 19 304 + 304 X 5 + ^ * 2 

= 20 849, from which should be subtracted the moment of wheel 1 
.about the right support. This moment of wheel 1 is 10 X 124 = 
1 240, and the moment about L 6 of all loads on the truss is 20 849 
- 1 240 = 19 609. The left reaction is: 

ft = 19 609 * 120 = 163.4 
M 3 = 163.4 X 3 X 20 - (5 848 - 10 X 64) 
= 4 596 000 pound-feet. 

In the case of L t , the reactions and the moments for the two 
positions are: 

For wheel 12, ft = 16364 -s- 120 = 136.4 

M 4 = 136.4 X 4 X 20 - 6 708 
= 4 204 000 pound-feet. 

For wheel 13, ft = (16 364 + 5 X 284 + ^-j^) + 120 = 148.4 
M t = 148.4 X 4 X 20 - 7 668 
= 4 204 000 pound-feet, 

which shows that each wheel gives the same moment, and also that 
the moment is greater than that at L 2 , the corresponding point on the 
left-hand side of the center of the truss. As L 2 is the center of 
moments of UJJ 2 , then, if the center of moments for U t U 5 falls at 
L 4 (that is, if LJJ 5 acts), the stress in U 4 U b will be greater than the 
stress in ?7 X ?7 2 when wheel 6 is at L 2 . Of course, if the engine came 
on the truss from the left, UjU 2 would receive the same stress that 
UJJ 5 now receives. According to the shears, LJJ 5 always acts, and 
therefore the center of moments for UJJ 5 does fall at L t . 



Stresses In a Pratt Truss 





The various moments are written 
in order, as such action will facili- 
tate the remainder of the computa- 
M, = 2 690 
M 2 = 4 100 
M 3 = 4 596 
M 4 = 4 204 

The chord stresses are now found 
to be: 

LL L L - l 2 69 l 107 5 

q 10 



O iO 

t^ O 




1 1 

cq q 
1 1 


4 100 


o 03 

1 1 

c^i -^ 


1 1 

U>U, 25 164.0 
r rr 4 596 

t/2C/3 ~ 25 
II U 4 2 4 16S - 



oq q o 
1 + 

(M O 


t/ 4 (/ 5 2 . [68.. 

L 2 L 3 = - C7,L 2 = - ( - 164.2) = + 164.2 
L 3 L 4 = -U t U s = -(- 168.2) = + 168.2 

When the load comes on from the 
left, the stresses in U 1 U 2 and L^ 
will be - 168.2 and + 168.2 respec- 
tively, which are the maximum live- 
load stresses for these members. 
Instead of placing the values of 
the stresses on a truss outline, they 
are sometimes put in tabular form, 
as in Table VII. 
48. Impact Stresses. WTien an 
engine is at rest on a bridge, the 
stresses in the members are the 
same as those computed for that 
loading. Wlien the loads move 
across the bridge at any speed, the 
vibrations and the shocks produced 
by the counterweights in the drivers 
and by other causes create stresses 
in the various members in excess 
of those computed by aid of the 
engine diagram. The excess stresses 


00 I- O 
C\ -* O 


10 q 


<* q cq 
+ + 1 

q oq 


p : 


+ 1 

00 t~ 
1 1 


2 d 
1 1 

<M r- 

I i 

S : 

co g o 

q cq 
oo co 



q <N q 
1 I 

<M O 

1 I 




Minimum. . 



are designated as impact stresses. This term, however, is mislead- 
ing to a certain extent, as causes other than the impact or pounding 
of the engine wheels help to produce the stresses referred to. 

It is a well-known fact capable of mathematical demonstration, 
that a load, if suddenly applied, will produce a stress equal to twice 
that which it will produce as a static load; also, that as the ratio of 
the weight of the load to the weight of the structure increases, the 
vibrations produced by the impact will be less. These two facts are 
the basis of most of the empirical formulae for impact stresses; and 
empirical formulse are used to obtain these stresses, as the existing 
conditions and producing causes are not such as to make them sus- 
ceptible of mathematical treatment. The result of experiments 
on actual bridges under the effect of passing engines and trains, have 
been the basis of many formulse. One of these is : 

where / = Impact stress in the member; 

S = Live-load stress in the member caused by the engine 

load when at rest; 
L = Length of that part of the bridge which is loaded when 

the stress S is produced; and 
300 = A constant. value derived from experiments. 

This formula was proposed by C. C. Schneider in 1887, and is 
given in the "Transactions" of the American Society of Civil En- 
gineers, Vol. 34, page 331. While it does not take into considera- 
tion the relative weights of the bridge and the live-load loads, this 
formula does make allowance for the time it takes to produce the 
stress, by introducing L, the distance over which the engine passes 
before causing the stress S. It is seen that the smaller the distance L, 
the greater will be the impact stress for any given value of S. When 
L becomes exceedingly small, the effect would be that of a suddenly 
applied load, and the impact stress would equal the stress S. Table 

VIII gives the values of j- -~ , which is called the impact co- 

efficient, for different values of L. Values not given may be inter- 

For example, by consulting Fig. 92, which gives the position of 
the engines for the maximum live-load stress in V \L 2 , it is seen that 
93 feet (the distance from* wheel 1 to the right support) is the loaded 




Values of the Impact Coefficient 







L + 300 

L + 300 

L + 300 

L + 300 

L + 300 





































































































Oci 7 

















1 3O 
















length. From Table VII, it is seen that the stress in U^ produced 
by this loading is +118.6; and from Table VIII, the impact coefficient 
for 93 feet is found to be 0.763. The impact stress is now computed : 

/ = 0.763 X 118.6 = +90.6. 
The maximum stress in U t L 2 is now : 

Dead-load = + 38.4 
Live-load = +118.6 
Impact = + 90.6 

Maximum = +247.6 

Table IX gives the necessary information for computing the 
impact stresses, and also gives the impact stresses corresponding to 
the maximum live-load stresses in the members of the truss of Article 



Impact Stresses in a Pratt Truss 







L + 300 

L U, 




125 .2 4 ft. of uniform load on truss 

U,L r 
U 2 L 2 
U 3 L 3 

+ 65.6 
- 50.5 
- 25.1 




+ 58.4 
- 41.3 
- 21.6 

See succeeding text. 
Same as for U 2 L 3 
Same as for U 3 L 4 

U,L 2 
U 2 L 3 
U 3 L, 

+ 118.6 
+ 64.7 
+ 32.0 




+ 90.6 

+ 52.7 
+ 27.6 

Wheel 16 at L R 
Wheel 11 is 4 ft. from L 6 
Wheel 9 at L . 







j Wheel 13 at L 4 . 
/ 5 ft. of uniform load on bridge. 
/ 15 ft. of uniform load on bridge. 
\ Wheel 1 off bridge. 

L L 3 
L 2 L 3 

+ 107.5 
+ 168.2 



+ 78.2 
+ 121.8 

Same loading as for L U l 
Same loading as for U 2 U 3 

In the case of U 1 L V it should be noted that only the wheels 10 to 
16 inclusive cause the stress (see Fig. 94), and that the loaded 
length is the distance from wheel 10 to wheel 16. 

Some specifications do not call for impact stresses. The unit- 
stresses in these specifications are made low, and the sections designed 
are large enough to withstand the additional stresses due to impact. 
In cases where the impact stresses are required, they must be con- 
sidered in computing the maximum and minimum stresses. 

49. Snow=Load Stresses. In some localities the snowfall is 
considerable, and its weight should be taken into account in com- 
puting stresses. This should be done by considering it as an addi- 
tional dead load of 15 pounds per square foot of floor surface for every 
foot of snowfall. As it covers the. entire floor surface, the stresses 
will be proportional to the dead-load stresses. Also it is evident that 
the snow load should not be taken into account in railroad bridges 
unless they have solid floors, as most of it falls through the open 
spaces between ties and stringers. 

As an example, let it be required to determine the snow-load 
stresses in a member of a highway bridge, the dead-load stress in the 
member being +84.0, the dead panel load being 12 000 pounds, and 



the snow be'rig H feet deep on the roadway, which is 14 feet wide. 
The snow panel load is: 

|- (14 X 15 X li X 20) = 3 150 pounds. 

In the above equation, 14 is the width of roadway; 15 is the weight 
in pounds of one square foot of snow one foot deep; and 20 is the 
length of one panel. One-half of the weight of snow must be taken, 
as half is carried by each truss. The snow-load stress is then : 

X 84.0- +22.1. 

In like manner, all snow-load stresses can be computed. 

Most of the standard specifications which have been published 
do not specify snow loads; and in fact it is not customary to include 
the snow load in any designs except those for bridges in extreme 
northern latitudes. It is hardly probable that the greatest load will 
come upon a country bridge when it is covered with snow. Also, 
in cities, the sidewalks are cleaned of snow; and so is the roadway 
if the city is of large size. 


50. Top Lateral System Through=Bridges. The unit-loads for 
this system are given in Article 26. Common practice is to take 
150 pounds per linear foot of top chord, the end-post being con- 
sidered part of the top chord in this computation. 

In many of the longer-span modern bridges, the diagonals of 
this system are designed to take either tension or compression ; but in 
the majority of the shorter spans, 200 feet and under, while generally 
consisting of angles or other stiff shapes, they are designed to take 
tension only. The verticals or top lateral struts take compression. 
This combination of tension diagonals and compression verticals 
makes the so-called Pratt system of webbing; and indeed the lateral 
systems, both top and bottom, are Pratt trusses in a horizontal posi- 
tion. Fig. 95 shows the side elevation of the truss of Article 47, and 
also the top and bottom laterals. The diagonals shown in full lines 
act when the wind is right, and those shown by dotted lines act 
when the wind is left. Wind right indicates that the wind is blow- 
ing from the right hand when a person stands facing the righ< end of 




the bridge. Wind left indicates that 

5 the wind blows from a person's left 

when standing as above described. 

The wind load of 150 pounds is di- 
vided between the two trusses, this 
being exact enough for practical pur- 
poses; for, by actual experiment, the 
< difference between the readings of 
g wind-pressure gauges placed at points 
o opposite each other in the top chords 
| of a through-bridge was only from 8 to 
^ 10 per cent. 

The problem, then, is one of a deck 
5 Pratt truss with a dead panel load of 
| 150 X 20 - 3.0 divided between the 
| two chords. Fig. 96 show T s the distri- 
^ bution of loads and the reaction, it 
* being considered that the portal brac- 
| ings and the end -posts (see Fig. 95) 
are stiff enough to distribute the 
g reaction equally between the bearing 
-f' / | >" I points L , L ', L 6 , L 9 '. Each panel 

5 load is indicated by an arrow, and 
d 3 is equal to 3.0 -r- 2 = 1.5. The re- 

S action at each of the points L , L ', 
| L 6 , and L 6 ' is 10x1.5 +4 = 3.75. 
g The truss being symmetrical, the 

6 stresses in like members on each side 
S of the center will be the same. The 
shears in the top system are: 

g F x = +2 X 3.75 - 2 X 1.5 = +4.5 

Fa-a = +2 X 3.75 - 3 X 1.5 = +3.0 
F 2 = +2 X 3.75 - 4 X 1.5 = +1.5 

and the secant ^ is (17 2 + 20 2 )* + 17 
= 1.544. The stresses in the diag- 
onals are: 


UiU 9 = +1.544 
U 2 'U a = +1.544 

X 4.5 = +6.95 
X 1.5 = +2.32. 




The vertical U 2 f U 2 = 3.0; and by passing a section 6 6 around 
U t f , the stress in U 3 f U 3 is found to be - 1.5. 

In obtaining the chord stresses in this system, the case is the 
same as if the reactions were applied at C7/ and U 5 ', as the portal and 
end-posts are not in the same plane as the lateral system. The tan- 
gent method is the simplest to use in this ca'se. The tangent is 
20 -r- 17 = 1.176, and the stresses (see Fig. 95) are: 

USU,' = - 4.5 X 1.176 = -5.29 

U 2 'U 3 ' = -(4.5 + 1.5) X 1.176 = -7.06 

U,U 2 = 

U 2 U 3 = -USUS = +5.29 

Fig. 97 is a diagram with the stresses caused by wind right and 

U e 


W.L. - 5.29 W.L. - 7.O6 

Fig. 97. Wind Stress Diagram of Pratt Truss of Fig. 95. 

wind left indicated thereon. The stresses for wind left can easily 
be written by inspection. 

51. Bottom Lateral Bracing, Through=Bridges. Fig. 95 shows 
the lower lateral system with the panel points loaded with the fixed 
or dead wind load. In this case it is all taken as acting on one side, 
it being assumed that the floor system protects the leeward truss. 
The problem then becomes that of determining the stresses in a deck 
Pratt truss of 6 panels of 20 feet each, the height being 1 7 feet. When 
wind is right, the members shown by broken lines in Fig. 95 do not act. 

The fixed wind load (Article 26) is 150 pounds per linear foot of 
chord. The panel load will be the same as before, 3.0, but all will 
be on one chord. The shears are : 

V, = 2* X 3.0 = +7.5 

Fa-a = +7.5 

F 2 == +7.5 - 3.0 = +4.5 

Fb-b = +4.5 

V a = +7.5 - 2 X 3.0 = +1.5 




The secant being 1.544, as previously computed, the web 
stresses are: 

L 'L l = +7.5 X 1.544 = +11.60 L/L, = -7.5 
L/L 2 = +4.5 X 1-544 = + 6.95 L 2 'L, = -4.5 
L 2 'L 3 = +1.5 X 1.544 = + 2.32 L 3 'L 3 = -3.0 

The stress in L 3 'L 3 is determined by passing section c c and 
resolving the vertical forces at L 3 f (see Fig. 95). 

By using the tangent method, the chord stresses are computed 
as follows: 

L/L/ = 
L,'L 3 ' = 
L,L 2 . 
L 2 L 3 - 

-7.5 X 1.176 = -8.82 
-(7.5 + 4.5) X 1.176 = -14.12' 
-(7.5 + 4.5 + 1.5) X 1.176 = -15.88 
-L/L/ = -(-8.82) = +8.82 


-(.- 14.12) = +14.12 

The wind load acting on the train is 450 pounds per linear foot. 
It is evident that the train may cover the span either partially or 
entirely, and therefore its action on the lower lateral system is the 
same as if it were stressed by a live load of 450 pounds per linear foot 
of truss. 

The live panel load is 450 X 20 = 9.0. The maximum live- 
load reaction is 5 X 9.0 +- 2 = 22.5, and the positive live-load shears 

V, = +22.5 

7 2 = (1 + 2 + 3 + 4) ^ = + 15.0 
V 3 = (1 + 2 + 3 ) -jr- = +9.0 

It is unnecessary to go further than the center, as only the maximum 
stresses are required in the members: The web stresses are com- 
puted as given below : 

L/L, = -22.5 
L 2 'L 2 = -15.0 
L 3 'L 3 = - 9.0 

The maximum chord stresses due to this load of 450 pounds per 
linear foot of train, occur when the train covers the entire span; and 
they are directly proportional to the stresses produced by the fixed 
load, in the same ratio as the live panel load is to the fixed panel load. 

This ratio is }- = 3.0. The chord stresses, therefore, are: 

L O 'L! = +22.5 X 1.544 = +34.75 
L/L 2 = +15.0 X 1.544 = +23.15 
L 2 'L 3 = + 9.0 X 1.544 = +13.91 




L 'L/ = 

- 8.82 X 3 = 
-14.12 X 3 = 
-15.88 X 3 = 
+ 8.82 X 3 = 
+ 14.12 X 3 = 

4 ~~f 

L/L/ - 
L 2 'L 3 ' = 
L,L 2 = 
+ 26.46 

+ 42.36 

X t 



Table X, Article 53, 
gives the stresses in the 
top and bottom lateral 
systems for wind right 
and wind left. 

52. Overturning Ef= 
feet of Wind on Truss. 
When the wind blows on 
the top chord, it tends to 
overturn the truss. As 
the truss is held down by 
its own weight, the action 
of the wind does not 
overturn it, but causes 
the dead-load reaction on 
the windward side to be 
less and that on the lee- 
ward side to increase by 
a like amount. The 

amount is V = -y X 

-,- . where 2w = the sum 

of all .the wind panel 
loads, h = the height of 
the truss, and b = the 
distance center to center 
of. trusses. The effect 
upon the leeward truss is 
the same as if two ver- 




tical loads, each equal to V and acting downward, were placed at the 
hips U 1 and U 5 (see Fig. 98). The effect on the windward truss is 
the same as if two vertical loads, each equal to V and acting upward, 
were placed at the hips C7/ and C7 6 '. 

The stresses in the leeward truss will now be worked out. The 
stresses in the windward truss are the same, but with opposite signs. 

10 X 1 5 2*1 
The truss is that of Article 47. Here V = X - = 11000. 

Fig. 99 shows the truss with the loads in the correct position, the 

Fig. 99. Truss under Wind Loads. 

reactions each being 11.00. V l = +11.00, and V 2 = +11.00 
11.00 = 0. The shears in the 2d, 3d, 4th, and 5th panels are also 
zero. As the shear in these panels is zero, the stress in the diagonals 
and vertical posts is zero X secant $ = zero. The stress in the hip 
verticals U 1 L 1 and U & L 5 is zero, as there are no loads at L l and L 5 . 
The stress in the end-post is -11.00 X 1.28 = -14.08. Taking 
the center of moments at U v the stress equation of L L l L t L 2 is : 
-I^a X 25 + 11.00 X 20 - 0; whence L t L 2 = +8.8. The stress 
in all the lower chord members will be found to be + 8.8. By 
summing the horizontal forces at the section a a, noting that, as 
f/jI/2 is zero, its component is also zero, there results: +L t L 2 + UJJ 2 
= 0; whence UJJ 2 = -LJL 2 = - (+8.80) = -8.80. This is also 
the stress in all members of the top chord. 

It is now seen that the overturning effect of the wind on the 
truss causes stresses only in the end-posts and chords. The wind on 
the lower chord causes no overturning effect, as it is transferred 
directly to the abutments. 

53. Overturning Effect of Wind on Train. The wind blowing 
upon the train tends to overturn it, and in so doing the pressure on 




the leeward stringer is increased and that on the windward stringer 
decreased by the same amount. This difference in pressures is 
transferred to the floor-beam and then to the panel points (see Fig. 
100), where its value is: 

L = 
where W 

W X (8.5 + a) 

Panel load due to wind 

on train; 

8.5= A constant established 
by the Specifications 
(see Article 26, p. 15); 
a = Distance from base of 
rail to center line of 
lower chord. It may 
be taken as 3 feet in 
most cases, as this is 
approximately the 
usual depth of floor). 
b = Distance center to cen- 
ter of trusses. 

For the case in hand, W = 20 
X 450 = 9 000. Therefore, 

9 000 X (8.5 

L = 



Illustrating Overturning Effect of 
Wind on Train. 

= 6 090 pounds. 

The action of the wind in tending to overturn the train is the 
same as if the truss were under a live panel loading of L, the panel 
load L acting upward on the windward and downward on the leeward 

The chord stresses due to this will be proportional to the dead- 
load stresses in the same ratio as this panel load L is to the dead panel 



For the truss of Article 47, this ratio is -- -- = 0.303, and 

20 000 

the chord stresses caused by the overturning effect of the wind on the 
train (see Table VII) are: 

U l U t = -64.0 X 0.303 = -19.39 
U 2 U 3 = -72.0 X 0.303 = -21.82 
L L, = L,L 2 = +40.0 X 0.303 = +12.12 
L 2 L 3 = +64.0 X 0.303 = +19.39 

The stress in U l L l is +6.09, the panel load at L v 
The maximum positive shears are: 

V v = ^ (1+2 + 3 + 4 + 5)= + 15.22 




V, = ^( 

1 + 2 + 3 + 4) = +10.15 

v,= ^( 

1 + 2 +^3) = +6.09 

y 4 = 12? ( 

1 + 2) = +3.05 

and the maximum web stresses are found to be: 

L U 1 = -15.22 X 1.28 = -19.50 

UJj t = +10.15 X 1.28 = +13.00 

U 2 L 3 = + 6.09 X 1.28 = + 7.80 

U 3 L 4 = + 3.05 X 1.28 = + 3.91 

U 2 L 2 = -6.09 
C7 3 L 3 = -3.05 

It is unnecessary to compute the shears further than one panel past 
the middle of the span, as only the maximum stresses are usually 

The wind stresses from various causes are grouped together and 
given in Table X. 

From Table X it is seen that large wind stresses occur in some 
of the members. Most specifications require that the stresses due to 
wind shall be neglected in the design unless they exceed 25 per cent 
of the sum of the dead-load and live-load stresses. 

The subject of wind stresses does not ordinarily receive the con- 
sideration it should have; in fact, it appears to be common practice, 
in the case of spans up to 200 feet, to neglect the action of the wind in 
all members of the bridge except the top and bottom lateral diagonals, 
the top struts, the portal, and the bending 
effect in the end-post. For the last two 
effects mentioned, see the next succeeding 

54. Portals and Sway Bracing. One- 
half of the wind on the top chord is trans- 
ferred to the hips U l 'U l and U 6 'U 6 . From 
there it is carried to the abutments by 
means of the portal bracing and the end- 
posts. Various styles of portal bracing are 
in use, but few are so easily analyzed and 
constructed as that of Fig. 101. This form 

Fig. 101. Style of Portal 
Bracing in Common Use on 
Spans up to 250 Feet. 




Wind Stresses in Pratt Truss 



L, V , 

U t L., 






Wind Right 

on Truss 



on Train 



+ 7.80 +3.91 

+ 6.09 



Wind Left 

on Truss 

+ 14.08 

on Train 

+ 19.50 


-7.80! -3.91 


+ 6.09 

+ 3.05 


+ Stress 

+ 33.58 

+ 13.00 

+ 7.80 

+ 3.91 

+ 6.09 

+ 6.09 

+ 3.05 


- Stress 










L,L n , 

L*L 3 


U,U 3 


1 . 

Wind Right 

+ 8.82 

+ 14.12 

+ 5.29 

+ 26.46 

+ 42 . 36 

Wind Left 

- 8.82 



- 5.29 

- 7.06 




Overturning Truss 
Wind Right 

+ 8.80 

+ 8.80 

+ 8.80 - 8.80 

- 8.80 

Wind Left 

- 8.80 

- 8.80 

- 8.80 

+ 8.80 

+ 8.80 

Overturning Train 
Wind Right 

+ 12.12 

+ 12.12 

+ 19.39 



Wind Left 



-19.39 ; +19.39 

+ 21.82 


+ Stress 

+ 20.92 

+ 56.20 

+ 84 . 67 

+ 28.19 

+ 35.91 


- Stress 








U t 'U, 

U,'U 3 

U t 'U, 

U 3 'U 3 


L,'L 2 


Wind Right 
on Truss 

+ 6 . 95 

+ 2.32 



+ 1 1 . 60 

+ 6.95 

+ 2.32 

on Train 

+ 34.75 

+ 23.15 

+ 13.91 

Wind Left 

on Truss 



on Train 


+ 6.95 

+ 2.32 



+ 46.35 

+ 30 ..10 

+ 16.23 

The stresses -in Z/ L,, L,f L,, and L 3 ' L 3 are not given in the above table. These 
members are the floor-beams, and the small stress due to wind Is neglected In their 



of portal is at present being used almost universally on all spans up 
to 250 feet. 

Let it be required to analyze a portal of this form, all the dis- 
tances being as indicated in Fig. 101 ; and let : 

w = Wind panel load of upper chord; 
m' = Number of panels in upper chord; 

P = (m' - 1) w; 

V = j (P + w} + w | A; 


H l = H 2 = { (P + w) + w \ + 2. 

The stress in BC, the center of moments being at D, is: 

. ^ = (P + w) a + HJ. = _ \ (P + W } + H J_\ 
a 1 2 a ) 

The stress in AB, the center of moments being at E, is: 

g,._ + j-it_ + . +tfl J, 

a a 

For the stress in BD, the center of moments is taken at C, and 
the perpendicular distance c to BD is determined. The stress in 
BD, then, is: 

<J -LW *1 

2 ~ 

The stress in BE is: 


It must be remembered that h v is not the height of the truss, 
but is the length of the end-post from L to U v 

For the truss of Article 47, w 1.5; m f = 4; and P = 4.5. 
The value h t - (20 2 + 25 2 )^ = 32.0 feet. The distance a must be 
so chosen that BD will not interfere with engines or other traffic 
which passes through the bridge. It will be assumed as 5 feet in 

this case. Then V = (4.5 + 1.5 + 1.5) ~ = 14.08; and H l = H 2 

= -^r = 3.75; whence, 

SBC = - (e.O + 3.75 X } = - 26.25 




SAB = 1.5 + 3.75 X - = +21.75 

The dis-tance BD = ^BC + CD 2 = ^8.5* -f- 5.0 2 = = 9.85. 
Then, from similar triangles DCS and DEC, is obtained the 
proportion : 


CD' BD' 

= +3.75 X 

4 . o 

= +27.90 

-3.75 X ~ = -27.90 

-4 . o 

When the wind blows from the other side, the stresses in the 
diagonals are reversed, and those in the top are transposed. The 
members shown by broken lines take no stress. When the wind 
blows, the end-posts tend to bend as shown in Fig. 102. This is with- 

Fig. 102. Illustrating Tendency 
of End-Posts to Bend under 
Wind Load. 

Fig. 103. Bending Tendency 

when Knd-Posts are Fixed 

at Lower End. 

stood by the cross-section of the post at the points E and D. The 
bending moment caused at these points by the wind is H l X I and 
H 2 X I. For the truss under consideration, 

M D = M E = 3.75 X 27 X 12 = 1 215 000 Ib.-ins. 

If the posts are fixed at the lower end, then they will tend to 
bend as shown in Fig. 103, the post resisting the bending at two 
points D and d. The section at each point withstands in this case 
only half of the moment just computed, or 1 215 000 -r- 2 - 607 500 
Ib.-ins. A further discussion of this will be given in Part II, on 
"Bridge Design." 




Various forms of sway bracing are used to connect the inter- 
mediate posts and thus stiffen the cross-section of the bridge at those 
points. The form of portal just given is often used, as is also the 
form shown in Fig. 104. Here h is the height of the truss. The 
braces BD are called knee-braces. Here w is the wind panel load of 

the top chord, and 

V = 


*.-*.- -r 

SBC = -(wo + HJ) -H a 

S B 'c> = + (w 



The stress in B 'B is the direct 
compression due to wind right or 
pig. KM. A Type of Portal and Sway left, and differs in accordance 

Bracing in Frequent Use. . , , . . i 

with the position of the top strut. 
There is also a bending moment at B' and B, which is: 

^MBI = MB = V'g + H 2 h. 

The bending moment at D and D' is equal to H 2 l or H 2 l -=- 2, 
according to whether or not the lower ends of the posts are fixed. 

The determination of the stresses for the truss of Article 47 is 
left to the student. 

When the wind is from the other side of the truss, the signs of 
the stresses in the knee-braces and the members C'B' and CB are 

55. Final Stresses. The class of stresses which go to make 
up the maximum or minimum for which the member is designed, is 
.determined by the specifications used. The dead-load and live-load 
stresses are always included, and then those due to impact and wind 
should be added if required. In computing the maximum stresses, 
the algebraic sum should always be used. In a large majority of 
cases, all stresses which go to make up the maximum have the 



same sign, but some exceptions have been noted, as in the middle 
vertical of a Pratt or Howe truss. The minimum stresses ure, with 
rare exceptions, obtained by combining stresses with signs of opposite 


56. Moments and Shears in Floor=Beams. In any bridge the 
floor-beam acts as a support for either the joists or stringers, and the 
moments and shears occurring in it are due to the loads which come 
on the joists or stringers. In a highway bridge the joists are spaced 
so closely that the load which they transmit to the floor-beams may 
be considered as uniformly distributed, providing the live load is a 
uniform load, in which case, 

, _ (2P L + Pn) X panel length in inches 


V = (2P L + P D ) * 2, 

where M = Maximum moment in pound-inches; 
V = Maximum shear; 
PL = Live panel load; 
PD = Weight of stringers and floor material in one panel. 

It will be seen that these formulae are those for the maximum 
moment and shear in a uniformly loaded beam, the total load being 
2P L + PD 

As an example, let it be required to determine the maximum 
moment and shear in the floor-beam of a highway bridge whose panels 
are 20 feet long, and trusses 16 feet center to center, the live load 
being 100 pounds per square foot of floor surface, the flooring weighing 
10 pounds per square foot, and there being 5 lines of joists weighing 
15 pounds per linear foot, and 2 lines of joists weighing 8 pounds per 
linear foot. 

Pi. = -- X 20 X 100 = 16 000 pounds. 

P D = 5 X 20 X 15 + 2 X 20 X 8 + 16 X 20 X 10 = 5 020 pounds. 


(2 X 16 000 + 5 020 ^ 20 X 12 
M = g 

= 1 110 600 pound-inches at center of floor-beam. 
V = (2 X 16 000 + 5 020) - 2 

= 18 510 pounds at ends of floor-beam. 




In the case of a single-track railroad bridge, there are only two 
stringers upon which the weight of the track, the engine, and the 
train is supported. These join the floor-beam at points equally 
distant from the center of same. The weight of the ties, rails, and 
fastenings is usually taken at 400 pounds per linear foot of one 
track. As regards the live load, the proposition reduces itself to 
placing the wheel loads so that the sum of the reactions of 
stringers in the adjacent panels will be a maximum on the floor- 
beam under consideration. This is discussed in Article 47, page 87 
(see Fig. 94). 

In determining the values of the maximum moment and shear in 
the floor-beam, the case is that of a beam symmetrically loaded with 
two equal concentrated loads. Each load is equal to the dead weight 
of one stringer, one-half the track weight in one panel, and the maxi- 
mum sum total of the reactions due to the wheel loads on the stringers 

in adjacent panels which, meet at 
that point. This latter quantity is 
called the floor-beam reaction. For 
a general arrangement of the loads, 
see Fig. 105. The distance a has 
become standard for single-track 
spans, and is 6 feet 6 inches. 

Let it be required to determine 
the maximum shears and moments 
in the floor-beam of the truss of 
Article 47. 

The weight of the stringer may be obtained by the formula of 
Table II, and is: 

Stringer = 20 (123.5 + 10 X 20) H- 2 = 3 200 pounds. 
The weight of one-half of the ties, rails, etc., in one panel is: 

i Track = (400 X 20) -H 2 = 4 000 pounds. 

The weight that comes from the engine wheels is given in Article 
47, page 87 (see Fig. 94), and is 65.55. Each load is therefore the 
sum of all the above weights, as follows: 

3 200 + 4 000 + 65 550 = 73 750. 

The maximum shear (see Fig. 105) is seen to be 73 750 pounds ; 
and the maximum moment occurs at C and D, and is: 




1 p 



^ Stringers ^ 

1 - 1 
Floor Beam 


c - 

. a.'6'.s 

5 17 Ft 

loC. of Truss? 

Fig. 105. Arrangement of Loads for Cal- 
culating Moments and Shears 
in Floor-Beams. 




M = 73 750 X (- " ) X 12 = 4 646 250 pound-inches. 

For any particular engine the floor-beam reactions for different 
length panels are easily tabulated for future reference. Table XI 
gives the floor-beam reactions for panel lengths from 10 to 24 feet 


Floor=Beam Reactions 
E 40 Loading 









41 000 




















61 000 








73 40 

In many cases it is desirable to keep the dead-load shears and 
moments separate from those of the live load; and this can easily be 

In neither of the above cases has the weight of the beam itself 
been taken into account. This 
should be done in the final design. 
The method of procedure is to 
compute the moment and shears 
as above; then make a provisional 
design of the beam. Next, com- 
pute the weight of the beam thus 
designed, and add the moments 
and shears caused by this weight 
to the other dead-load moments 
and shears ; then re-design the beam 
and compute its weight. If this 

last weight varies 10 per cent from the previous weight, another re- 
design should be made. The above proceeding belongs to Bridge 
Design, Part II, and will there be treated. 

57. Moments in Plate=Girders. Plate girders are of two classes 
namely (1) those which have the ties or floor laid directly upon the 
upper flanges of the girders; these are called deck plate-girder bridges; 

Fig. 106. Cross-Section of Deck Plate- 
Girder Railway Bridge. 




and (2) those in which the webs of the girders are connected with 
each other at intervals by floor-beams which in turn carry stringers 
or joists in exactly the same manner as in the floor system of a 
railroad or highway truss-bridge ; this latter type is called a through 
plate-girder bridge. Figs. 106 and 107 show cross-sections of deck 

and through plate- 
girder bridges re- 
spectively, for rail- 
way service. Fig. 

108 is a side view 
of a deck plate- 
girder bridge. Fig. 

109 is a longitudinal 
section of a through 
plate-girder rail- 
road bridge. The 

section is taken down the middle of the track. The bridge shown has 
5 panels. An odd number of panels should be chosen, as this does 
not bring a floor-beam at the center of the span, and hence the great 
moment which would then be caused is avoided. 

The analysis of the shears and moments of a through plate- 
girder is precisely the same as that for a truss bridge. The shear is 

Fig. 107. Cross-Section of Through Plate-Girder Railway 


*fl m \m &. 

m m m. 

m B8LJH 

m w BB 

1 M 

Fig. 108. Side View of Deck Plate-Girder Railway Bridge. 

constant between any two panel points as 0-1 or 1-2, etc., and the 
moments are computed for the points 1, 2, 3, and 4. ^ 

If wheel loads are used for moments, the relation that K = 

-Lmustbe. + , and that k = - (L + P) must be-, holds 

true when the loads are in correct position for maximum moments. 
Here m = the number of panels, and n = the panel under considera- 




tion and is to be reckoned from 
the left end ; in fact, all terms 
have the same value as men- 
tioned in Article 46. A careful 
review of Articles 44 and 46 
should enable the student to 
follow the example which will 
now be given. 

EXAMPLE. It is required 
to determine the moments at 
the points of floor-beam support 
for a 5-panel through plate- 
girder of 75-foot span. The 
live loading is Cooper's E 40. 

Dead-Load Moments. 
Through plate-girders, on ac- 
count of the heavy floor system 
and the fact that the floor sys- 
tem transfers its own weight 
and that of the live load to the 
girders as concentrated loads, 
are about 40 per cent heavier 
than deck plate-girder bridges 
of the same span. The weight 
of the entire span, therefore, is; 

1.4 X 75(123.5 + 10 X 75) = 
91 700 pounds. 

Part of this 91 700 pounds (the 
weight of the girders themselves) 
acts as a uniform load ; the re- 
mainder (the weight of the 
floor-beams and stringers) acts 
as concentrated loads at the 
points where the floor-beams 
join the web. Experience has 
shown that the weight of the 
floor for a single-track railroad 
system is about 400 pounds 





Position, Moments in a Through Plate-Girder 

-- c 




L + P 



^ - 34.4 






^j 2 = 38.4 





~ = 4QA 





* = 40.4 





^ = 39.0 






152 X 2 60 8 







172 X 2 68 8 



/ u 




152 X3 

1 Q 






172 X 3 103 2 

1 ^ 




1 1 fi 

192 X 3 

1 q 


1 ID 





129 X 4 1Q3 

1 S 

1 ro 










1 lo . O 



152 X 4 






152 X 4 



IZI . O 






per linear foot. The weight of the stringers and floor-beams for 
this bridge is therefore 75 X 400 = 30 000 pounds, and 91 700 - 
30 000 = 61 700 pounds acts as a uniform load. This 61 700 pounds 
is distributed over two girders, and so gives 61 700 -r- (2X75) = say, 
412 pounds per linear foot of one girder. 

The dead load which is concentrated at each panel point is that 
due to the weight of the steel floor and the weight of ties, rails, and 
fastenings. It is, for one girder, 

15 X (400 + 400) -f- 2 = 6 000 pounds. 




! 5 II 

3* I* 


S S^^-E 

^ l?fli 

< aj 

^ 5? 

a in 

3 u 




The dead-load moments are now computed by the methods of 
Strength of Materials, and are found to be : 

M = M t = 0; 

M t = M 4 = +4 390 000 pound-inches; 

M 2 = M s = +6 580 000 pound-inches. 

Live-Load Moments. The positions of the wheels for maximum 
moments are now determined (see Table XII). 

The computations for the reactions are best arranged in tabular 
form. Table XIII gives the values. 

Reactions for a Through Plate-Girder 

IH * 
2 \a<o 

gLp * 

1 i 3 

1 | 4 
1 5 

~~2 4 

3 | 6 
3 7 



R = (6 708 
R = (7 668 
72 = (8 728 

+ 192 X 4) - 75 
+ 212 X 4 - 10 X 78) * 75 
+ 232 X 4 - 10 X 83 - 20 X 75) - 75 


R = (4 632 

+ 152 X 7) -H 75 


R = (4 632 
72 = (5 848 

+ 152 X 6) -J- 75 
+ 172 X 3) H- 75 


4 6 

4 7 
4 8 

R = (2 851 
72 = (3 496 
72 = (4 632 

+ 129 X 4) -=- 75 
+ 142 X 4) + 75 
+ 152 X 2) H- 75 


The live-load moments are computed as follows: 

I Wheel 3, M = 99 .7 X 15 - 230 = 1 265 1)00 pound-feet. 
Wheel 4, M =103.2 X 15 - 20' X 5 - 20 X 10 = 1247000 
Wheel 5, M = 97.8 X 15 - 20 X 5 - 20 X 10 = 1 167000 

Point 2 Wheel 4, M = 76.0 X 30 - 
[Wheel 6, M = 73 

Point 3 

480 = 1 800 000 pound-feet. 
X 45 - 1 640 = 1 785 000 pound-feet. 

"[Wheel 7, M = 84.9 X 45 - 2 155 = 1 665 000 pound-feet, 
f Wheel 6, M = 44.9 X 60 - 1 640 = 1 054 000 pound-feet. 
54.2 X 60 - 2 155 = 1 097 000 pound-feet. 

Point 4^ Wheel 7, M 

I Wheel 8, M = 65.4 X 60 - 2 851 =1 073 000 pound-feet. 

The above values show that the greatest live-load moments are: 

M at Point 1, by wheel 3 = 1 265 000 pound-feet 
Mat " 2," " 4=1800000 " " 
Mat " 3, " " 4 = 1800000 " " 
Mat " 4, " " 3 - 1265000 " " 



The last two values are obtained when the load comes on the bridge 
from the left. Inspection of the results obtained at points 3 anti 4 
when the load comes on from the right, shows that they are con- 
siderably smaller than the results obtained at their symmetrical 
points 1 and 2, and therefore it was not necessary to determine the 
moments for any points to the right of the center. This is true of all 
girder spans, deck or through. 

The method of procedure when the girder is a deck plate-girder 
is the same as that just illustrated, except that in the computation 
of the dead-load moments there is no concentration of certain por- 
tions of the dead load, the weight of the girders themselves being a 
uniform load, as is also the weight of the ties and rails or, if it be a 
highway bridge, the floor-joists which run transversely. Highway 
spans are seldom built of deck plate-girders, it being preferable to 
use the through girders, as then the girders themselves serve as a rail- 
ing and keep the traffic confined to the roadway. The girder span is 
usually divided into ten equal divisions, the points of division being 
called the tenth-points. The shears and moments are computed for 
the center point and those points which lie to the left of the center. 
After the values are computed, they are laid off as ordinates, with the 
corresponding tenth-points as abscissae. A curve is then drawn 
through their upper ends, and the curve of maximum shears or 
moments is the result. To get the maximum shear or moment at 
any point other than a tenth-point, the ordinate is scaled at the 
desired point. 

EXAMPLE. Let it be required to determine the maximum 
moments at the tenth-points of a 100-foot-span deck plate-girder. 

Dead-Load Moments. The weight of steel in the span is 100 
(123.5 + 10 X 100) = 112 350 pounds, and the weight of the track 
is 400 X 100 = 40 000 pounds, making a total of 152 350 pounds, 
or 152 350 * (2 X 100) = say, 762 pounds per linear foot per girder. 
The dead -load moments are now determined according to the methods 
of Strength of Materials, and are : 

M, = 342 800 pound-feet 

M 2 = 609400 " " 

A/ 3 = 799 840 " " 

M 4 = 914 100 " " 

M 5 = 952 200 ' ' " 
Live-Load Moments. The determination of the wheel load 




positions is made by the use of the formulae K = ( L) and 

w m 

k = (L -f P) ; only, in this case, n is the number of divisions 


from the left support to the section, and m is the number of divisions 
into which the girder is divided. 

The determination of the wheel positions is given in Table XIV. 

Wheel Positions, Moments in Deck Plate-Girder 






L + P 







258 X 0.1 = 25.8 
261 X 0.1 = 26.1 
254 X 0.1 = 25.4 
242 X 0.1 = 24.2 
240 X 0.1 = 24.0 
230 X 0.1 = 23.0 








232 X 0.2 = 46.4 
245 X 0.2 = 49.0 
258 X 0.2 = 51.6 
261 X 0.2 = 52.2 











232 X 0.3 = 69.6 
232 X 0.3 = 69.6 
245 X 0.3 = 73.5 
261 X 0.3 = 78.3 










212 X 0.4 = 84.8 
232 X 0.4 = 92.8 
245 X 0.4 = 98.0 
258 X 0.4 = 103.2 
261 X 0.4 = 104.4 










232 X 0.5 = 116.0 
232 X 0.5 = 116.0 
245 X 0.5 = 127.5 
258 X 0.5 = 129.0 
274 X 0.5 = 137.0 
244 X 0.5 = 122.0 
234 X 0.5 = 117.0 
224 X 0.5 = 112.0 
234 X 0.5 = 117.0 







While many wheels on point 1 satisfy the condition, the greatest 
moment will occur when one of the large drivers is at the point, and 
it is therefore unnecessary to examine the point for other wheels. 
The same is true at the center point, 5, the maximum occurring under 
one of the heavy driver wheels. The reactions and the computations 
for the same are given in Table XV. 




Reactions for a Deck Plate-Girder 







R = (12041 + 5 X 258) -.100 
R = (12 599 + 4 X 261) H- 100 
R = (11 984 + 4 X 254) -5- 100 
R = (11 334 + 4 X 234 -T- 2 X 4 a - 2) - 100 




R = 12041 H-100 
R = (12 041 + 5 X 258) -r- 100 




fl = 10 816 - 100 

R = 12 041 -5- 100 





R = (8 728 + 4 X 232) - 100 
fl = (10 816 + 4 X 245) -f- 100 
fl = (12 041 + 4 X 258) -f- 100 

130 . 73 



# = (8 728 + 8 X 232) H- 100 
R = 12041 -100 
R =(12041 + 5X 258) * 100 
R = (13 904 + 2 X 274) H- 100 
R =(11 334+ 5X234 + 2X 5 2 - 2) - 100 
R = (9 514 + 10 X 214 + 2 X 10 2 - 2) - 100 
R = (7794 + 15 X 194 + 2 X 15 2 - 2) H- 100 


Maximum Moments in a Deck Plate-Girder 










M = 

M = 
M = 
M = 

133.31 X 10 
136.43 X 10 
130.00 X 10 
122.86 X 10 

- 80 
- 5 X 20 
- 5 X 20 
- 5 X 20 

1 253 100 
1 200 000 
1 128 000 



J!f = 

M = 

120.41 X 20 
133.31 X 20 

- 230 
- 480 

2 178 200 



M = 

M = 

108.16 X 30 
120.41 X 30 

- 480 
- 830 

2 764 800 
2 782 300 



M = 

M = 

M = 

96.56 X 40 
117.96 X 40 
130.73 X40 

- 830 
- 11 640 
- 2155 

3 032 400 
3 078 400 
3 074 200 



M = 
M = 
M = 
M = 
M = 
A/ = 
M = 

105.84 X 50 
120.41 X 50 
133.31 X 50 
144.52 X 50 
125.29 X 50 
117.54 X 50 
109.29 X 50 

- 2 155 
- 2851 
- 3496 
- 4072 
- 3068 
- 2658 
- 2248 

3 137 000 
3 169 500 
3 169 500 
3 154 000 
3 196 500 
3 219 000 
3 216 500 



Table XVI gives the computations of the live-load moments at 
the tenth-points, the final results being in pound-feet. 

Whenever any loads were off the left end of the bridge, the lines 
7 to 16 of the engine diagram were used (Fig. 85). For example, with 
wheel 10 at 5, wheel 1 would be off the left end. By looking in the 
second space of line 8, there is found the quantity 13 904, which is 
the moment of wheels 2 to 18 inclusive about a point directly 
under wheel 18. Just to the right of the vertical line through 
wheel 18, is the value 284, which is the weight of wheels 1 to 18 in- 
clusive; but this must be decreased by 10, the weight of wheel 1, as 
that wheel is off the span. As wheel 18 is 2 feet from the right end 
of the girder, the moment about the point is 13 904 + 274 X 2. By 
looking in the second space of line 16, the value 4 072 is found. This 
is the value of the moment of loads 2 to 9 inclusive about a point 
directly under wheel 10, and must be subtracted from the moment 
of the reaction in order to get the moment at 5 for this loading. 
See Articles 21 and 47 for further information regarding the use of 
the values in lines 7 to 16 of the engine diagram. 

By the help of differential calculus it can be proved that the 
greatest possible moment does not occur at the middle of a beam loaded 
either with concentrated loads or with concentrated loads followed by 
a uniform load, but it occurs under the load nearest the middle of the 
beam when the loads are so placed that the middle of the beam is half 
way between the center of gravity of all the loads and the nearest load. 

The wheel which produces this greatest moment is the same one 
which produces the maximum moment at the middle of the beam. 
The exact solution of this problem involves the use of quadratic 
equations, but for all practical purposes the following rule will 
suffice : 

Place the loading so that the wheel which produces the maximum 
moment at the middle of the beam is^at that point. Find the distance of the 
center of gravity of all the loads from the right end. Move the loads so that 
the middle of the beam is half way between the center of gravity as found 
above and the load which produced the maximum moment at the middle 
of the beam. Find the moment under that load, with the loads in the position 
just mentioned. 

For the case in hand, wheel 12 at 5 gives the maximum moment. 
The moment at the right end of the span, wheel 12 being at 5, is: 

9 514 + 10 X 214 + 2 X~10 2 -7-2 = 11 754 000 pound-feet. 



The center o; gravity is - = 50.2 feet from the right sup- 


port, or 0.2 foot to the left of the center of the girder. Now place 
whee' 12 one-tenth of a foot to the right of the center, and deter- 
mine the moment under it. The reaction will be : 

R = (9 514 + 9.9 X 214 + 2 X 9~9 2 + 2) H- 100 = 117.306; 

M = 117.306 X 50. 1 - 2 658 = 3 219 306 pound-feet. 

In this particular case the difference between the greatest 
moment possible and the greatest moment at the middle is not 
sufficient to warrant the extra labor involved in computing it. In 
general it may be said that if the greatest moment possible occurs within 
six inches of the middle of the beam, it is not necessary to compute it, 
the moment at point 5 being taken. 

58. Shears in Plate=Girders. In the case of through plate- 
girders, the maximum live-load shears are determined by placing the 

wheels in such a position that Q = - - G is + , and q = 
(G + P)is-. 

In these equations, m is the number of panels into which the 
span is divided; and the other quantities are the same as given in 
Article 45, which should now be reviewed. 

For example, let it be required to determine the dead and live 
load shears in the through plate-girder of Article 57, p. 111. 

The weight of one girder = 61 700 -T- 2 = 30 850 Ibs. 
" "4 the floor = 5 X 6 000 = 30 000 Ibs. 

Total weight on one girder = 60 850 Ibs. 

The dead-load shears are then computed by the methods given in 
Strength of Materials, and are given as follows, it being remembered 
that the concentrated load which comes at the end is one-half a panel 
load, or 3000 pounds: ' 

T = 60 850 H- 2 = 30 475 Ibs. = end shear; 

or ocri 

F, = 30475 - 3000 - =21 305 Ibs.; 


or* o c f\ 

V 2 = 30 475 - 3 000 - 6 000 - 2 X * = 9 135 Ibs. 


V 3 = 30 475 - 3 000 - 2 X 6 000 - 2 X - = 3 135 Ibs. 




where V = the shear at the end ; V l = the shear just to the left of 
point 1; V., = the shear just to the left of point 2; V 3 = the shear 
just to the right of point 2 ; and Vc = the shear at the middle of the 

The determination of the wheel load position for maximum 
live-load shears is given in Table XVII. By comparing the formulae 
Q and K, it will be seen that for the first panel Q = K, and q = k, 
as n = 1. The position of wheel loads for maximum moments at 
point 1 is the same as for maximum shear in the first panel. Accord- 
ing to Table XII, wheels 3; 4, and 5 at point 1 all give maximum 
shears in the first panel. In this case, as in previous ones, only the 
shear for the first position of the loading found- for any particular 
point will be determined, as the difference between this and the other 
cases is too small to warrant the additional labor necessary in com- 
puting them. It is evidently unnecessary to go past panel 3, as only 
the maximum shears are required. 

Wheel Positions, Shears in a Through Plate-Girder 














See Table XII, and 

text above this table 




142 -5- 5 = 28.4 









152 -s- 5 = 30.4 








152 -5- 5 = 30.4 






116 Hh- 5 = 23.2 









116 H- 5 = 23.2 ! 20 



For wheel 3 at point 1, the left reaction (see Table XIII) is 99.7. 
That portion of wheels 1 and 2 which is transferred to point is 
230 -r- 15 = 15.33; and the shear in the first panel, therefore, is: 

V l = 99.70 - 15.33 = +84.37. 
For wheel 2 at point 2, the left reaction is : 

R = (3 496 + 142 X 5) -^ 75 = 56.10; 

F 2 = 56.10 --5- = + 50.67. 



A computation with wheel 3 at point 2 will give a shear only 560 
pounds greater, which difference would not influence the design to 
any appreciable extent. 

For wheel 2 at point 3, the left reaction is: 

R = (2 155 + 116 X 1) -5- 75 = 30.30; 

> 3 = 30.3 - ~= +24.97. 
1 o 

When the girder is a deck one, the computation of the dead 
shears is very much simplified, as' all of the load is uniform. 

Fig. 110. Beam under Wheel Loads Followed by Uniform Load. 

Let it be required to determine the dead and live load shears 
at the tenth-points of the deck plate-girder of Article 57. 
The total weight of one girder and track is 76 175 Ibs. 

V = 76 175 -*- 2 = +38 088 pounds. 

V, = 38088 - 7 ~^ = +30 470 pounds. 

V 2 = 38 088 - X 76 175 = +22 450 pounds. 
V 3 = 38 088 - |0 X 76 175 = + 15 230 pounds. 

V t = 38 088 - ~ X 76 175 =+ 7 618 pounds. 

^ s = 0. 

The position of the wheel loads to produce the maximum shear 
cannot be determined by the same relation as that used in structures 
which have a system of floor-beams and stringers, for here not a 
portion, but all of the load to the left of the section, must be sub- 
tracted from the left reaction in order to give the shear. 



The correct relation f6r the wheel load position will now be 

Let Fig. 1 10 represent a beam of span I loaded with a series of 
wheel loads followed by a uniform load. Let P equal the weight of 
the first wheel, W equal the weight of all the loads, and g the distance 
from the center of gravity of all of the loads to the right abutment. 
The distance between the first and second wheel centers is a, and the 
first wheel is at the section b - b at a distance x { from the left support. 

;? ' W 9-- 
R, =- r , 


F' b _ h = R,' - (loads to left of section) = ^X 

Now, assume that the loads move forward the distance a. The 
wheel 2 will be at section b-b, and Fig. Ill will represent the 
position of the loads. Then, 

W (g + a). 
, z 


F" b _ b = R," - P 

_.W(g + a) . 

- + - P. 

It is now evident that in order to get the greatest shear at section 
b-b, wheel 2 must be placed at the section whenever F" b _ b .is 
greater than F' b -b- Then, 

F" b _ h > F' b _ b ; 

Wg ,Wa Wg 

_ ---- r _ P >__. 

Now, canceling out the term ^-, which appears on both sides of the 


equation, there results: 

For the engine under consideration, a = 8 feet, and P = 10 000 
pounds, and the equation reduces to: 




which is to say that when the load on the girder is greater than \\ 
times the span, then wheel 2 should be placed at the section in order 
to give the maximum shear. 

For loading E 40, the following is true : 

For all' sections up to and including the center of all spans, place 
wheel 2 at the section to give the maximum shear. 

In Fig. Ill it is immaterial whether or not any additional loads 
come on the span at the right end when the loads move forward the 

> a I 

It S 




r /"K JL s~*\ ^ 

) I 



1 u x ' ^ 




Fig. 111. Beam of Fig. 110 with Loads Moved Forward. 

distance a, as they would only tend to increase the left reaction and 
therefore the shear F" b _ b . If the relation deduced is true for the 
case when no extra loads come on at the right end, it will be true 
when they do. 

The live-load shears at the left end and at the tenth-points, 
wheel 2 being at the section in all cases, are computed from the gen- 
eral formula, which is: 

V = R - I P, 
in which, 

R = Left reaction; 

I P = All loads to left of section, and is equal to 10 000 pounds for 
all sections except the end of the girder. 

The computations and results can be conveniently placed in 
tabular form, and are given in Table XVIII. 

In order to illustrate the use of the relation W> 1 I, let point 

3 in the above span be taken. Place wheel 2 at point 3; then, as 
wheels 1 to 13 are on the girder, the total weight W is 212. As I = 
100, \\l = 125. Therefore, as 212 is greater than 125, wheel 2 is 
the correct wheel. 




Maximum Shears in a Deck Plate-Girder 









(13904 + 4X274)^100 



Wheel 18, 4 ft. from rt. end 






Wheel 16, 5 ft. from rt. end 
Wheel 15 at right end 


( 7 668 + 4X212) -100 




Wheel 13, 4 ft. from rt. end 


( 5848 + 4X172)^-100 




Wheel 11, 4 ft. from rt. end 


( 4632 + 2Xl52)-hlOO 




Wheel 10, 2 ft. from rt. end 

The curves of maximum live-load moments and shears are shown 
in Fig. 112. They should always be drawn. From them the shear 
or moment at any desired section 
can be determined. For exam- 
ple, let it be desired to determine 
the maximum live-load shear and 
moment at a point 24 feet from 
the left end of the girder. By 
drawing the ordinate, shown by 
a broken line in Fig. 112, and 
scaling, the following values are 
found : 

V M = 88 000 pounds; 

M, 4 = 2 440 000 pound-feet. 

A similar set of curves for 
the dead-load shears and moments 
should be made. The set for the 
deck plate-girder in hand isshown 
in Fig. 113. These are easily 
constructed by laying off the max- 
imum values of the shear at the 
end, and the maximum value of 
the moment at the center. The 
shear curve is a straight line from the end to the center, while the 
moment curve is a parabola from the center to the end. 

The stresses in the lateral systems of plate-girders are computed 
in a manner the same as that employed for the lateral systems of 
trusses, the unit-load being taken according to the specifications used. 












a ; 



















/ - 


























, | 


A 2 

/ u 




















. g 






Fig. 112. Curves of Maximum Live-Load 
Moments and Shears. 




59. Stresses in Plate=Girders. The stresses in plate-girders 
are treated in the Instruction Paper on Steel Construction, Part IV, 
pages 251 to 263, and the student is referred to this treatise for infor- 
mation regarding this subject. 

The stress in the flange is seen to depend upon the distance from 
center of gravity to center of gravity. This distance, in turn, depends 

Fig. 113. Curves of Dead-Load Shears and Moments in a 100-Foot Span Deck Plate-Girder. 

upon the depth of the girder. Certain approximate rules have been 
proposed in order to determine this, but the following formula will 
give the width of the web plate in accordance with best modern 
practice : 


d = 

0.005 I + 0.543' 

in which 

d = Width of the web plate, in the even inch; 
I = Span, in feet. 

For example, let it be required to determine the width of the 
web plate of a plate-girder of 80-foot span center to center of end 


0.005 X 80 + 0.543 0.94 

= 85.2 (say 86) inches. 

If the resultant value had been 85 inches, the width would have been 
taken as either 84 or 86. The reason for this is that the wide plates 
kept in stock at the mills are usually the even inch in width and can 
therefore "be procured more quickly than if odd -inch widths were 
ordered, in which case the purchaser would be forced to wait until 




they were rolled often a period of several months. The distance 
back to back of flange angles, the so-called depth of girder, is one-half 
inch more than the width of the web. This is due to the fact that 
each pair of flange angles extend one-fourth inch beyond the edge of 
the web plate, so as to keep any small irregularities caused on the 
edge of the web plate by the rolling, from extending beyond the backs 
of the angles. 


1. Determine the maximum positive shears in the first six panels of 
a 9-panel 114-foot Pratt truss, the live panel load being 8.0. Use the exact 
and also the conventional method. 







+ 32 . 00 

+ 32 . 00 

V 2 


+ 24.54 

+ 24.90 

V 3 


+ 18.05 

+ 18.70 

V 4 


+ 12.48 

+ 13.35 

V 5 


+ 7.85 

+ 8.90 



+ 4.50 ' 

+ 5.34 

2. Find the maximum and minimum stresses in LJJ 2 and U 3 L 3 of an 
8-panel 160-foot through Warren truss. Height 20 ft.; dead panel load 10.00, 
all on lower chord; live panel load 12.00. 

ANSWER: In L,U 2 : d. 1., -28.00; 1.1., - 35. 30 and +1.68; 
max., -63.30;min.,"-26.32. In U 3 L 5 : d. 1., +16.80; 1.1., + 25. 20 
and -5.04;max., +42.00;min., + 11.76. 

3. In the truss of Problem 2, determine the maximum stress in L 2 L 3 
by the method of moments, and also by the tangent method. 

ANSWER: d. 1. = +67.50; 1.1. = +81.00; max. - +148.50. 

4. Determine the dead-load stresses in the members U 2 L 2 and L 4 U 6 
of a 9-panel 180-foot through Warren truss. Height is 24 feet; dead panel 
load is 10.0, one-third being at each panel point of the upper chord, and two- 
thirds being at each panel point of the lower chord. 

ANSWER: J7 2 L 2 = + 30 . 60 ; L,U, = - 1 . 80. 

5. Determine the stress in the counter of a through Howe truss of 8 
panels and 160-foot span. Height is 30 ft.; dead panel load, 9.6; live panel 
load; 11.5. 

ANSWER: 4.59. 




6. In the truss of Problem 5, determine the maximum and minimum 
stress in U 2 L 2 , L 2 U 3 , and L 3 C/ 4 . 


U t L t 

L 2 C/ 3 

L 3 U 4 

d. 1. 

+ 20 . 80 
+ 30.30 


+ 5.18 

- 5.76 


+ 15.10 
+ 19.36 

-43.20 -23.06 
-21.20 0.00 

Fig. 114. Deck Parabolic Bowstring Truss. 

7. In the deck parabolic bowstring truss of Fig. 114, determine the 
maximum stress in LjL 2 , L,C/ 2 , and U 3 L 3 . The dead panel load is 4.0, all 
on upper chord; and the live panel load, 20.0. 

ANSWER: Z^L, = +201.9; L,^ = +21.8; I7 3 L 3 = 33.6. 

Fig. 115. Through Bowstring Truss. 

8. In the through bowstring truss of Fig. 115, determine the maximum 
stress in f7,L 2 and L^U^, the dead panel load being 5.0, and the live panel 
load 15.0. 

ANSWER: U,L 2 = +33.50; 





9. Determine the maximum and minimum stresses in the members 
{/,/ f/,L 2 , UJLi 2 , and C7 3 L 3 of a 7-panel 175-foot through Pratt truss 30 feet 
high. Dead panel load is 10.0, all on lower chord; live panel load is 15.0. 



V t L+ 

C/ 2 L 2 

U 3 L 3 

d. 1. 
1. 1. 

+ 10.0 
+ 15.0 

+ 26.00 
+ 41.70 

- 2.78 

+ 6.42 




+ 25.0 
+ 10.0 

+ 67.70 
+ 23.22 


- 3.58 


10. Determine the maximum and minimum stresses in the members 
C/jWij, m 3 L 3 , U 2 L 2 , and m 2 U 2 of the deck Baltimore truss shown in Fig. 110. 
Dead panel load, 30 000 Ibs.; live panel load, 50 000 Ibs. One-third of dead 
panel load is applied at the lower ends of all the verticals. 


L\m t 

m 3 L 3 



d. 1. 

+ 190 '8 
+ 333 . 5 
- 15.1 

+ 84.8 
+ 191.5 
- 50.5 

+ 10.7 

+ 21.2 
+ 56.6 



+ 524 . 3 

+ 175.7 

+ 276.3 
+ 34.3 

- 99.3 

+ 56.6 
+ 21.2 

11. In the truss of Problem 10, determine the maximum stress in 
M 2 t/ 2 andL 3 L 4 . 

ANSWER: M 2 U 2 = -840.0; L 3 L 4 = +960.0. 

12. Determine the position of the wheel loads of Cooper's E40 loading 
to produce the maximum positive live-load shears in the panels of a 7-panel 
175-foot Pratt truss. 

ANSWER : L v wheel 4 ; L 2 , wheels 3 and 4 ; L a , wheel 3 ; L 4 , wheel 
3; L 5 , wheel 2; L 6 , wheel 2. 

13. Determine the maximum positive live-load shears for the truss 
of Problem 12. 

ANSWER: V, = 192.8; F 2 - 137.8; F 3 = 90.8; F 4 - 52. G; 
F 5 = 25.0; F = 6.8. 




14. Determine the position of the wheel 
loads of Cooper's E 40 loading to produce max- 
imum moments at the panel points of the truss 
of Problem 12. 

ANSWER; L v wheel 4; L 2 , wheel 7; 
L 3 , wheels 11 and 12; L t , wheels 13 and 

15. Determine the maximum moments at 
the panel points of the truss of Problem 12. 
Loading, Cooper's E 40. 

ANSWER: M l = 4820000; M 2 = 
7745000; M 3 = 9192000; M 4 = 
9 082 000, all in pound-feet. 

16. Compute the maximum live-load web 
| stresses in the truss of Problem 12, the height 
H being 32 feet. Loading, E. 40. 

17. Compute the maximum live-load chord 
73 stresses in the truss of Problem 12, the height 
| being 32 feet. Loading, E 40. 

18. Compute the impact stresses for all 

members of the truss of Problem 12. 

19. Determine the maximum live-load 

g 1 shears at the tenth-points of a 65-foot span 
deck plate-girder. Loading, E 40. 

ANSWER: F = 103.0; V l = 86; F 2 
= 69.7; F 3 = 54.5; V 4 = 40.8; F 5 = 

20. Compute the shear due to impact in 
the girder of Problem 19. 

ANSWER: F = 84.7; V l = 71.5; F 2 
= 58.8; F 3 = 47.0; F 4 = 35.6; F 5 = 

21. Compute the maximum live-load 
moments at the tenth-points of the girder of 
Problem 19. Loading, Cooper's E 40. 


















11 320 














1 .45' from center 




All moments are in thousands of pound-inches. 




60. General Economic Considerations. The prime considera- 
tion which influences the decision to build is cost. After the decision 
to build has been made, the problem is one of a purely engineering 
character, whereas in the first case it was one of either a political or 
an engineering character, or both. The engineering problem is an 
economic one, in which maximum benefits must be obtained at a 
minimum cost. 

A map of the proposed bridge site and the approaches, as well 
as of the country for a considerable distance up and down stream, 
should be made. This map should show the contours, the soundings, 
the borings, the high and low water-mark elevations, and the excep- 
tional flood line. On this map the bridge should be plotted in its 
proposed location and also in various others. In the case of each of 
these locations, various schemes taking into account different numbers 
of piers and spans should be considered. 

Several authors have attempted to present formulae having a 
more or less theoretical derivation and purporting to indicate the 
correct number of piers and spans for a minimum cost. The use of 
these formulae should not be encouraged, since they do not in any case 
give results close enough to serve for anything but a rough guide. 

The cost of abutments will vary somewhat with the location and 
the character of the approach. This variation is usually small, and 
ordinarily an approximate location of the abutments can be quickly 
made. As the number of abutments is in all cases constant, their 
effect upon the problem of the location of the bridge is small, the main 
proposition being that of the cost and the number of piers and spans. 

The cost of the piers will usually not be constant, those closer to 
the middle of the stream costing more on account of the depth of the 
water and the more difficult character of the foundation. Piers 

Copyright, 1908, by American School of Correspondence. 



should not be placed on a skew; neither should they be placed directly 
in the maximum line of action of the current. If a skew is unavoid- 
able, it should be as small as possible. The cost of piers should be 
ascertained by the most careful estimates. In the case of small 
bridges where there are only one or two piers, the matter is very simple, 
but with a considerable number of piers the problem becomes very 
complicated and requires weeks and sometimes months or years for 
its solution. 

The determination of the cost of the superstructure is a com- 
paratively simple matter. In certain instances the class of bridge is 
limited to some extent by the specifications. Cooper, in Article 2 of 
his "Specifications for Steel Railroad Bridges and Viaducts" (edition 
of 1906), gives the following: 

Types of Bridges for Various Spans 



Up to 20 feet 
20 to 75 " 
75 to 120 " 
120 to 150 " 
Over 150 " 

Rolled beams 
Riveted plate-girders 
Riveted plate- or lattice-girders 
Lattice or pin-connected trusses 
Pin-connected trusses 

One railroad expresses a preference for plate-girders for all spans 
from 20 to 115 feet; and for spans from there to 150 feet, riveted 

The question as to whether the bridge will be deck or through 
is one which is decided by the controlling influences of water-way, 
false work, time of erection, and extra cost of masonry. If the clear 
height required for the water-way is sufficiently small, the deck 
bridge should be chosen, as in this class the cost of false work is less, 
the time of erection is less, and the cost of masonry is less by an amount 
equal to the cross-section of the piers times the depth of the truss. 
Deck bridges also cost less than through bridges of equal span. 

The conditions permitting, girders should be used in preference 
to trusses. While for equal spans girders are heavier and therefore 
cost more, the steel work alone being considered, little or no false 
work is required, and the time of erection is much less than in the 



case of trusses. This makes the total cost of girder bridges less than 
those in which trusses are used. Another item in favor of girders is 
their great stiffness. 

While pin-connected bridges cost less and are easier to erect, 
their stiffness is not so great as that of riveted bridges, which cost 
more. The time required for the erection of riveted bridges is also 
greater than that for pin-connected bridges. This is on account of 
the great amount of time required to make the riveted connections. 
For long spans, say over 200 feet, it is necessary to use pin-connected 
bridges, as the extreme size of the connection plates prohibits the use 
of the riveted type. Also, it is unnecessary to use riveted long-span 
trusses to obtain stiffness, as the weight of the pin-connected bridges 
is so great when compared with the live load that sufficient stiffness 
is obtained. 

The cost of spans of different lengths and character may be 
obtained directly from the bridge companies; or their weights may 
be computed from the formulae given in Article 20, p. 9 (Part I, 
"Bridge Analysis"), and multiplied by the unit price which your 
experience indicates is correct, thus giving the total cost. 

Evidently the solution of problems of this nature cannot be 
made within the limits of this text, but the following example will 
tend to indicate somewhat the manner of procedure in a problem of 
this kind. For example, if the length between abutments is 1 400 ft., 
the cost of each abutment is $12 000, and the cost of each pier is 
$15 000, then, if we have fourteen 100-foot plate-girder spans, each 
costing $4 300, and thirteen piers, the total cost will be $279 200. On 
the other hand, if nine piers and ten 140-foot truss spans, each cost- 
ing $9 200, are used, the cost will be $251 000, showing a balance of 
$28 200 in favor of the truss scheme. The live loading is E 50. 

61. Economic Proportions. The depth of girders is given in 
Article 59, Part I. 

In the case of trusses, the effect of an increase in the height 
is to increase the stresses in the web members and to decrease the 
stresses in the chord members. This variation does not affect the 
weights to any considerable extent; in fact, a variation of 20 per cent 
in the height will not affect the weight more than 2 or 3 per cent. 

The height of the bridge is usually fixed by some considerations 
which are in turn determined by the specifications. The height must 




Panel Lenqtl 
istoes feet. 

Doable Track 

be sufficient to clear whatever traffic will pass through. It should 
also be sufficient to prevent overturning on account of the wind 
pressure on the truss or on the traffic. In addition, the height of the 
bridge is influenced by the depth of the portal bracing. A deep 
portal bracing is desirable, in that it stiffens the trusses under the 
action of the wind and the vibration due to the passing traffic; but 
a deep portal bracing increases the height of the truss and therefore 

^ ft the bending in the 

end-posts due to 
the wind. Judg- 
ment on the part 
of the engineer 
should be used in 
order to determine 
the limiting height 
for securin gamax- 
imum amount of 
benefit as regards 
stiffness and a 
minimum amount 
of bad effect due 
to the bending in 
the end-posts. Fig. 
117, which gives 
the height for any 
given length of 
span, may be said 
to represent the best modern practice (1908). Variations of a foot 
or more from those given do not affect the weight to any appreciable 

The distance from center to center of trusses for highway 
bridges depends upon the width of the street or, if in the country, 
the width of the roadway. Streets, of course, vary in width in differ- 
ent localities, but country highway bridges usually have a roadway of 
from 14 to 16 feet in the clear. 

In the case of railroad bridges, the distance from center to center 
of trusses depends upon whether the track is straight or on a 
curve, and also upon whether the bridge is a deck or a through bridge. 

Single Track 

5 pan in fct. 

Fig. 11 

Curves Showing Relation between Height of Trusses 
n Double 
ay Bridg 

and Length of Span in Double- and Single-Track 



The actual amount varies in most cases, and is fixed by specification. 
Some specifications require that when the track is straight, the dis- 
tance from center to center of trusses shall be 17 feet; or that, in case 
one-twentieth of the span exceeds the 17 feet, then one-twentieth of 
the span shall be used. 

For deck plate-girders the common practice appears to be to 
space them as given below : 

Width of Plate-Girder Bridges for Various Spans 



Up to 65 feet 
65 to 80 " 
80 to 115 " 

6 feet 6 inches 
7 feet inches 
7 feet 6 inches 

For through plate-girders the spacing should be such that no 
part of the clearance diagram will touch any part of the girder. In 
case of double-track plate-girders with one center girder, great care 
should be exercised in order that the center girder shall not be so deep 
nor have so wide a flange as to interfere with the clearance diagram 
(see Fig. 126). 

On account of the wind on a train which runs on track placed 
at the elevation of the top chord of deck bridges, the overturning 
effect is exceedingly great, and special care should be taken that the 
height and width are such as to prevent overturning. 

In through bridges the clearance must be such as to allow the 
clearance diagram to pass. Special attention should be paid to the 
knee-braces and also to the portal braces. When the bridge is on a 
tangent, the spacing of the trusses is a comparatively simple matter, 
being just sufficient for the clearance diagram; but on curves, allow- 
ance must be made for the tilt of the diagram due to the super- 
elevation of the outer rail, and also allowance must be made for the 
fact that the length of the cars between trucks forms a chord to the 
curve, and as such the middle ordinance must be taken into account. 
It is also necessary to allow for that part of the car which projects 
over the trucks, as this will extend beyond the outer rail by an amount 
greater than one-half the width of the clearance diagram. (See 
Figs. 119 and 120.) 




; -^ 








; . 



JC 50 



\ H 







d Hi? 

* c i; 

Z . -s-J 








. Sbl 


S ^ - ^ 


o B-^Z 

11 ~ |C 





- Sals 
g ls*a 

i ^2o a 



""" o^ 




=* ?"/_ 






M 9 -,9 





> =i=^ 


3 :?fs 


* S^^ 















gram (Straight Track) U 

raight tracks. When tr 
ucks. Also, on account 
'i inches is to be proYide 
structures of lesser heigl 


" ^~^^ 

" 5 si? 

cO a> c"o o 



o lill 








Q g dH 

Sl i 

| 725 
^ l? = 

1 iji 

3 $ 


: o 

2>U\~\ 7>i]U.2>^) 



. K 5 a 
% ^S3 
^ ^l" 

^ 5? " 


r i 








62. The Clearance Diagram. 
The clearance diagram is not 
supposed to represent the outline 
of the largest engine or car which 
may run over the line, but repre- 
sents the maximum amount of 
space which may be taken up by 
objects which are to be shipped 
over the line. For instance, the 
lower part of the clearance dia- 
gram may allow for snow-plow 
or ballast distributors, and the 
upper part may take into account 
the passage of such material as 
carloads of lumber, piles, or tele- 
graph poles. The standard clear- 
ance diagram of the Lehigh 
Valley Railroad is given in Fig. 
118. This diagram is for the 
clearance on straight track only. 
On curves, the diagram tilts as 
show r n in Fig. 119, and to allow for this tilting the Lehigh Valley 
Railroad requires 2.V inches additional clearance on the inside of 
curves for each inch of elevation of the outer rail. In addition to this 
tilting effect, the clearance should also be increased on account of the 

Fig. 119. Clearance Diagram on Curves, 
Showing Tilting. 

Fig. 120. Standard Car on Curve, Showing Necessity for Wider Spacing of Trusses. 

length of the cars and their projection over the outer and inner rails. 
Fig. 120 shows a standard car according to the specifications of the 
American Railway Engineering & Maintenance of Way Association, 
in such a position on a single-track span as to show the effect of the 
curve upon the widening of the spacing, center to center of trusses. 



This car is 80 feet long, 60 feet between centers of trucks, and is as 
wide as the clearance diagram, 14 feet for single track. It is evident 
that the trusses cannot be spaced so as to interfere with the clear- 
ance line of the body of the car and its projecting ends. These 
clearance lines are represented as broken lines in Fig. 120, and are 
marked c-c. Note that the center of the track is seldom in the center 
of the floor-beam. Also, it is evident that the sharper the curve, 
the greater the required distance between trusses, and accordingly 
the greater the floor-beams in length. This varies the moment in the 
different floor-beams and therefore makes them more costly. The 
stringers, also, are more costly, on account of the fact that their ends 
are skewed. On account of the eccentricity of the track, one truss 
takes more of the load than the other, and therefore the trusses are 
not the same a fact which further increases the cost. 

From the above it is seen that almost all conditions incident to 
the building of a bridge on a curve tend to increase the cost; and 
hence a fundamental principle of bridge engineering: Avoid build- 
ing bridges on curves. 

63. Weights and Loadings. For the weight of steel in any 
particular span, and for the loading required for any particular class 
of bridge, see Articles 20 to 23, Part I. The weight of the ties and 
the rails and their fastenings is usually set by the specifications at 
400 pounds per linear foot of track. For highway bridges the weight 
of the wooden floor is usually taken at 4^ pounds per square foot of 
roadway for every inch in thickness of floor. 

Highway bridges are divided into different classes according to 
their loadings (see Cooper's Specifications). The decision as to the 
class to be employed depends somewhat upon the distance to the 
nearest bridge across the same stream. In case the nearest bridge 
is only a few miles away and is of heavy construction, it is not actually 
necessary to construct a heavy bridge at the proposed site, the heavier 
traffic being required to pass over the other bridge. In case a heavy 
bridge is not in the neighborhood, then one shpuld be constructed at 
the proposed site. If the proposed site is on a road connecting adja- 
cent towns of large size, then a heavy bridge should be constructed 
and provision made for future interurban traffic, even if none is at 
that time in view, since it will be more economical to do this than to 
erect a new bridge in the future. 













<D ^ 




y So: 

L. X. 


CQh li- 

W] O (D 


^ s.1 

. en 







O ^ 



O - 


^ o 


In the case of railroad b'ridges, new ones are nearly always con- 
structed to carry the heaviest main line engines. These are usually 
of a class corresponding to Cooper's E 40 or E 50. In some localities 
branch-line bridges are built for the same live loadings; but in the 
majority of cases the branch-line bridges consist of the old bridges 
from the main line. 

64. Specifications. For any particular bridge the specifica- 
tions are either written by the engineer in charge, or some of the very 
excellent general specifications which are on the market in printed 
form are used. Some railroads use these general specifications 
with the addition of certain clauses which are desired by the chief 
or bridge engineer. The principal differences in these general 
specifications are in regard to the allowance for impact. 

Whenever highway design is mentioned in this text, it is to be in 
accordance with Cooper's Highway Specifications (edition of 1901). 
Wherever plate-girder design is given, it is in accordance with Cooper's 
Railway Specifications (edition of 1906); and wherever truss design 
is given, it is in accordance with the general specifications of the 
American Railway Engineering & Maintenance of Way Association 
(second edition, 1906). 

65. Stress Sheet. Before the sections are designed, the com- 
puter makes a skeleton outline of the truss, and on this places the 
dead -load and live-load stresses, and, in case the wind should be 
considered, the wind-load stresses. This is sent to the designer. The 
designer determines the various sections, and also the moments and 
shears in the stringers and floor-beams. These are placed on a sheet 
usually 17 by 23 inches. This is called a stress sheet. This sheet is 
now given to the draftsman, who makes a shop drawing. The stress 
sheets for railroad bridges are usually more elaborate than those for 
highway bridges. Plate I is the stress sheet of a highway bridge; 
and Plate II (Article 78) and Plate III (Article 93) are examples of 
the best modern practice in the making of plate-girder and truss- 
bridge stress sheets. 

66. Floor System. Perhaps no part of bridge design is better 
standardized than the construction of the open steel floors for 
railroad bridges. The stringers are usually placed 6 feet 6 inches 
apart, and consist of small plate-girders, or, if the panel length is 
.short, of one or more I-beams. I-beams are economical in regard to 




Safe Spans for I Beams 

(Based on unit-stress of 10 000 Ibs. per square inch in extreme fibre) 



1 - 



Safe span C to C of Bearings 

Safe span C to C of Bearings 

1 Beam 
per rail 

2 Beams 
per rail 

3 Beams 
per rail 

1 Beam 
per rail 

2 Beams 
per rail 

3 Beams 
per rail 

9 in. 
9 " 
9 " 
10 " 
10 " 
10 " 
12 " 
12 " 
12 " 
12 " 
15 " 
15 " 
15 " 
15 " 
15 " 
18 " 
18 " 
18 " 
18 " 
18 " 
20 " 
20 " 
20 " 
20 " 
20 " 
24 " 
24 " 
24 " 
24 " 
24 " 



. 890 
1 063 

1 180 
1 277 
1 453 
1 650 
2 112 









t. 3i 







18 ' 

12 ' 


t. 9 in. 
9 " 
6 " 
9 ' 
9 ' 

6 ' 
5 ' 

7 ' 
7 ' 


b. 9 ii 




I t 















t. Oin 
' 9 




' ' 

4 ' 










t. 6 in. 
' 3 " 
' 6 " 
9 " 
9 " 
3 " 
6 " 
6 " 
6 " 
6 " 
6 " 
9 " 
6 " 
9 ' 
6 ' 
9 ' 
3 ' 
3 ' 

3 ft. in. 
" 6 " 

" 9 " 
" " 
' 9 " 


15 ' 

18 ' 

15 ' 

16 ' 

Ml ' 
20 ' 

2(1 ' 


22 ' 


6 ' 

3 ' 
9 ' 
6 " 
9 " 
6 " 

8 ' 


HI ' 

9 ' 



3 " 
3 " 

9 " 

3 " 
9 " 
6 " 
6 " 
6 " 
9 " 
3 " 
6 " 

11 ' 
12 ' 


14 ' 

14 ' 


16 ' 


16 ' 

ir i 



6 ' 
6 ' 

24 " 6 ' 

23 " ' 
24 " ' 

12 ' 
11 ' 

3 " 

9 " 
3 " 
' 6 " 





3 " 

' " 
' 9 " 
' 6 " 
' " 
' 3 " 

26 " 6 ' 
28 " ' 
29 " 3 ' 
29 " 9 ' 
30 " 9 ' 
31 ' ' 
32 " ' 

13 ' 6 " 
14 ' " 
14 ' 6 " 
14 ' 9 " 
15 ' " 
15 ' 3 " 


20 ' 
21 ' 

21 ' 
22 ' 
22 ' 




Ills 1111 

0' O" 0" O O" (Cf tO" <0" 

nj tf t <T tf OjfuSjS 

o OOOO o o o o 

0000 0000 

o oooo oooo 

' t i sj" s?~ sj' K" si" n 


8 555 9 565 

- 8' -S'-S'-S'- 9 1 -5- 6'- 5'- 

first cost, but are disadvantageous on account of the eccentric con- 
nections which necessitate heavy brackets to resist part of their re- 
action. They are also somewhat undesirable on account of the fact 
that, the ties deflecting, most of the load is carried by the inner 
I-beam. However, I-beams for stringers and for short-span bridges 
(see Fig. 121) are much used in present practice, and give good re- 
sults. Figs. 121 to 127 show the standard open floor sections of the 
I^ehigh Valley Railroad. Table XIX gives the required number of 
I-beams, together with their weight, which are to be used for short- 
span bridges or as stringers in panels of given length. 

Solid floors consist of angles and plates, channels and plates, or 
other shapes. They extend transversely across the bridge from truss 




i. s'-n" il 

to truss, the lower 
chords, in case of 
truss bridges, be- 
ing made heavy 
enough to act as 
girders as well as 
tension members. 
Figs. 128 to 130 
show sections of 
solid floors. The 
ballast is laid di- 
rectly upon these 
solid floors, which 
are first covered 
with a good damp- 
proof paint. The 
floors should also 
be supplied with 
good drainage fa- 


Lb ^'V 

l e i ^ H? 

t rs=^_U_ 8x9x10' 

T ^ i r" * 

,-. c c T 

Fig. 121. Sectional View Showing Open-Floor Construction of 
Railroad Bridge of Short Span, Single Track. I-Beams 
used for Stringers. Lehigh Valley Standard. 



t [Hrr~~ii~- :rr5. 8 * 9 " lo '-' 



nF- 1 v 


-g," - 


Concrete is 
sometimes laid di- 
rectly upon the steel floor, and the ballast put upon this concrete, 

Fig. 122. Section of Open-Floor Construction of Deck-Girder 

and Truss Bridge, Single Track. Lehigh Valley Standard. 

Plate-Girders used for Stringers. 

Fig. 123. Floor Construction of a Through-Girder Bridge, Single Track. 
Lehigh Valley Standard. 




which has previously had a 
"layer of some good waterproof- 
ing applied on its upper sur- 

67. Practical Considera= 
tions. The possibilities of the 
rolling mill and the various 
shops of a bridge company, 
such as the drafting room, 
forge, foundry, templet shop, 
assembling shop, and riveting 
and finishing shop, and also 
the shipping and erect *r.g facil- 
ities, should be well known in 
order to make the most eco- 
nomical use of them. This 
requisite knowledge comes 
only from experience. The 
best way to obtain this experi- 
ence without being actually 
employed in the shops, is to go 
into the shops every chance 
that presents itself, keep your 
eyes and ears open, and ask all 
the questions you can. The 
use to be made of handbooks 
of the various steel manu- 
facturers is given in Part I 
of "Steel Construction," and 
should be thoroughly studied 
before going further. Some 
one of these handbooks is in- 
dispensable to persons design- 
ing steel structures. That of 
the Carnegie Steel Company 
(edition 1903) is one of the 
best, and will be frequently 
referred to in the present text. 



Copies may be procured from the Carnegie Steel Company, Frick 
Building, Pittsburg, Pa. The usual price to students is 50 cents, 
to others $2.00. 


68. The Masonry Plan. In some cases the general dimensions 
of the masonry are limited ; such a case, for example, would occur in 
the crossing of a street or narrow waterway. Here the length of the 
span and the distance above the street or the surface of the water, 

Drain Piper I 1 . 

Fig. 129. Solid-Floor Construction of Plates and Angles. 

are the limited dimensions. The span and under-clearance may be 
unlimited, as in the case of a country stream crossed by a roadway 
which is a considerable distance above the surface of the water. 
The term unlimited is not here used in its exact meaning, as the span 
in this case is really limited by the cost, which rapidly increases with 
the length of the span. 

In some cases, as when the engineer is in a bridge office, the 
masonry plans are sent in by the railroad. In such cases many of the 
limited dimensions are fixed. The most usual dimensions to be fixed 
are the elevation of base of rail, the elevation and size of the bridge 
seat, and the length of the span under coping. These limit the 
depth of the girder, or the depth of the floor if it be a through 
girder, and also limit the length of the bearing plates at the end. 
Fig. 131, the masonry plan of a road crossing, shows in general what 
can be expected. All the dimensions usually fixed are given, and those 
marked x and y may or may not be, but x should never be less than 
3 feet. 

69. Determination of the Class. As before mentioned, the 
deck plate-girder should be used if possible, since its cost is less. 
There are some cases, however such as track elevation in cities 
where the additional cost required to elevate the track so as to use a 



deck plate-girder will more than balance 
the saving in its favor. In such cases the 
through plate-girder is used. 

The case whose design is under con- 
sideration will be taken similar to that of 
Fig. 131, and the span will therefore be a 
deck one. 

70. Determination of the Span, Cen= 
ter to Center. Fig. 132 shows the various 
spans namely, under coping, center to cen- 
ter of end bearings, and over all. The span 
under coping is that span from under cop- 
ing to under coping lines of the abutments, 
and is so chosen as to give the required dis- 
tance between the abutments at their base. 
The span center to center is equal to the 
span under coping plus the length of one 
bearing plate. The span over all is the 
extreme length of the girder. The length 
of the bearing plate is influenced by the 
width of the bridge seat, and also by the 
maximum reaction of the girder. The 
length should seldom be greater than 18 
inches and never greater than 2 feet, as 
the deflection of the girder will cause a 
great amount of the weight to come on the 
inner edge of the bearing plate and also on 
the masonry, in which case the masonry is 
liable to fail at that point and the bearing 
plates are over-stressed. 

Cast-steel bearings are now almost 
universally used. They decrease the height 
of the masonry, and distribute the pressure 
more evenly and for a greater distance 
over the bridge seat. When these castings 
are used, the bearing area between them 
and the girder may be made quite small, 
thus doing away to a great extent with the 




deteriorating effect due to the deflection of the girder as mentioned 
above. Fig. 133 shows the end of a girder equipped with a cast- 
steel pedestal. Table XX gives the length of the bearing on the 

asg of Rail- 

elevation j; 

[/Center Line ot Road 

ipan under Copinq 

= ep'-o- 


Elevation 756. 






Fig. 131. Masonry Plan of a Road Crossing. 

masonry for various spans, Cooper's E '40 loading being used and 
cast-steel pedestals being employed. 

Length of Masonry Bearings 



15 to 24 f( 
25 to 44 
45 to 69 
70 to 79 
80 to 89 
90 to 115 


12 inches 

As an example, let it be required to determine the span center 



^- Masonry Plate Masonry Bearino'rF= i 
r Span Under Copim) M 


\ Soon Center to Center Bearinq ""^ 

, ' \ Span Over AH / 

Fig. 132. Diagram Showing Various Spans Considered in Bridge Construction. 

loading being E 40. From Table is seen that the length of 
the masonry l>caring will 1x3 21 inches, and therefore the span center 




to center of bearings will be GO + 2 X ($ X 1 ft. 9 in.) = 61 feet 
9 inches. 

In Articles 71 to 77 the above girder will be designed ; and also 
such information as is of importance regarding the subject-matter 
of each article will be treated. The dead- and live-load shears are 
computed by the methods of Part I, and are given in Fig. 134. 

71. Ties and Guard=Rails. The length of ties for single-track 
bridges is 10 feet. 
For double-track 
bridges the 
length is in most 
cases the same. 
In some double- 
track bridges, 
however, either 
each tie or every 
third tie extends 
entirely across 
the bridge. In 
other cases every 
third tie on one 
track extends to 
the opposite 
track, thus act- 
ing as a support 
for the foot-walk 
which is laid 

Upon them. It Fig 133. End of a Girder Equipped with a Cast-Steel Pedestal. 

is the best prac- 
tice to limit the length of the ties on double-track bridges to 10 feet, 
since, if they extend into the opposite track in any way whatsoever, 
unnecessary expense is incurred whenever repairs or renewals are 
made, because both tracks must necessarily be disturbed to some 

The size of the ties varies with the weight of the engines and the 
spacing of the stringers or girders on which they rest. They are 
usually sawed to size instead of hewn, and the following sizes may be 
purchased on the open market namely, 6 by 8, 7 by 9, 8 by 9, 9 by 10, 






Fig. 134. Shear and Moment Diagram. 



and 10 by 12 inches. Larger sizes may be obtained on special order. 

The elevation blocks (see Fig. 127) should be of length to suit the 
width of the cover-plates and the spacing of the supports. They are 
usually made of the best quality of white oak, since the cost of renewal 
is great enough to demand that they be made of material as permanent 
as possible. 

The guard-rails should be placed in accordance with the specifica- 
tions (see Articles 13 and 14). Some railroads specify that the guard- 
rails shall be in 24-foot lengths unless the bridge is shorter than 24 
feet, in which case one length of timber should be used. For method 
of connection and other details, consult Figs. 121 to 127. The 
guard-rails and the ties are usually made of Georgia long-leaf yellow 
pine, prime inspection. Other wood, such as chestnut, cedar, and 
oak, may be used. 

In addition to the wooden guard-rail, a steel guard-rail usually 
consisting of railroad rails is placed within about 8 inches of the 
track rail. 

In designing ties, the problem is that of a simple beam symmet- 
rically loaded with two equal concentrated loads, the weight of the 
rail and tie itself usually being 
neglected. For the case in hand, 
which is that of a deck plate- 
girder, loading E 40, the con- 
centrated load for which the tie 
must be designed is, according 
to Specifications (Article 23, 3d 

rar^ Q 3*^ rrmnrlc A^nrv-Hr,. Fig. 135. Distribution of Loading on Ties Of 
partj, O 666 pounds. According Deck plate-Girder Bridge. 

to Article 23, 100 000 pounds is 

on four wheels. This gives 25 000 pounds on one wheel, and ac- 
cording to Article 15, one-third of this, or 8 333 pounds, will come 
on one tie. Fig. 135 shows the condition of the loading, the space 
center to center of rail being taken as 4 feet 10 inches. Some 
designers take this distance as 5 feet; but as the standard rail head 
is about 2 inches, and the standard gauge 4 feet 8 inches, the distance 
here taken seems to be the more logical one. 

The formula to be used in the design of this beam is that given 

in "Strength of Materials," and is M - In this case M = W X 




8 333 = 83 330 pound-inches. In the above formula, / = bd*+ 12, 
and c = d ~- 2, and therefore - = -. Substituting the value 

of the moment in the above formula, and solving for S, there results' 

499 980 

For a 6 by 8-inch tie, the unit-stress w T ould then be : 
S = 4 6 99 ^ = 1 310 pounds. 

If a 7 by 9-inch tie is used, the unit-stress is found to be 880 pounds. 

Since according to Article 15 of the Specifications, the unit-stress 

cannot be greater than 1 000 per 
square inch, it is necessary to use 
a 7 by 9-inch tie. If the engine 
loading had been E 50, the mo- 
ment would have been 100000 
pound-inches, and then the stress 
in a 7 by 9-inch tie would be 1 060 
pounds per square inch, and the 
stress in an 8 by 9-inch tie would 
be 930 pounds, which would neces- 
sitate the use of the latter. 

The guard-rails on this bridge 
will be placed according to the 
Lehigh Valley standard, and hence 
their inner face will be 4 feet 1 
inches from the center of the 

Elevation blocks will not be 
required, as the bridge is on a 
72. The Web. The economic depth of the web, according to 

Article 59, Part I, will be: 

61 ft. 9 in. 

Fig. 136. End Rivets Transferring Shear 
to Web. 

Depth = 


0.005 X 61 ft. 9 in. + 0.543 
The depth might be taken as 72 inches, but 74 inches will be decided 
upon, as this will decrease the area of the flange and also will not 
affect the total weight to any great extent- The unit-stress for shear 




is 9 000 pounds per square inch (see 
Specifications, Articles 40 and 41). 

The maximum shear in the girder 
occurs at the end, where it is 117 800 
pounds. The area required for the web 
is then 117800 -r- 9000 = 13.09 square 
inches, and the required thickness is 13.09 
-T- 74 = 0.177 inch. This latter value 
cannot be used, since, on account of Ar- 
ticle 82 of the Specifications, no material 
less than f inch can be used. The web 
plate will therefore be taken as 74 in. by 
f in. in size. 

Some engineers insist that the net 
section of the web should be considered. 
Consider Fig. 136, the shear being trans- 
ferred to the web by the end rivets. The 
web will not tend to shear along the 
section B-B, in which case the rivet-holes 
should be subtracted; but it will shear 
along section A- A, a section which is 
unaffected by the rivet-holes. The web 
splice should come at one of the stiffeners, 
and will therefore be considered in Arti- 
cle 76. 

73. The Flanges. This portion of 
the girder is usually built either of two 
angles or of two angles and one or more 
plates. In heavy girders where the flange 
areas are large, additional area is ob- 
tained by using side plates or side plates 
and four angles. Sometimes two chan- 
nels are used in the place of side plates 
and angles. Fig. 137 shows the different 
methods of constructing the top flanges 
of girders. The lower flanges are usually 
of the same construction. Fig. 137 b has 
the web extending beyond the upper sur- 




et or less 


faces of the upper flange angles. This is done in order that the 
ties may be dapped over it, and thus prevent the labor usually 
required for cutting holes in the lower face of the tie in order to 
allow for the projecting rivet-heads. Fig. 137 g is usually uneconom- 
ical, since the thinness of the channel web requires a great many 
rivets to sufficiently transmit the shear from the web to the flange, 
and also since the cover-plates must be very narrow. 

Specifications usually state that the flanges shall have at least 
one-half of the total flange area in the angles, or that the angles shall 
be the largest that are manufactured. The largest angles are not 
usually employed, since their thickness is greater than three-quarters 
of an inch and therefore the rivet-holes must be bored, not punched. 
The reason for this is that the depth of the rivet-hole is too great in 
proportion to its diameter, and on this account the dies used for 
punching frequently break. Also, the punching of such thick material 

injures the adjacent metal, which 
makes it undesirable. In reality 
the flange area of only the short- 
span girders is small enough to 
allow the flange area to be taken up 
by the angles. 

In choosing the thickness of the 
cover-plates, care should be taken 
so that the outer row of rivets will 
not come closer to the outer edge 
of the plate than eight times the 
thickness of the thinnest plate. In 

case eight times the thickness of the plate is greater than 5 inches, 
then 5 inches should be the limit. Also, the distance between the 
inner rows of rivets should not exceed thirty times the thickness of 
the thinnest plate. These limitations are placed by Article 77 of the 
Specifications, and Fig. 138 indicates their significance. 

The determination of the required flange area depends upon the 
distance between the centers of gravity of the flanges; and in order 
to determine this exactly, the area and composition of the flanges 
should be known. The above condition requires an approximate 
design to be made, the supposition being that the flanges consist of 
two angles and one or more plates as shown in Fig. 138. . 

Fig. 138. Diagram Showing Relation be- 
tween Thickness of Cover-Plates 
and Position of Rivets. 



The distance back to back of angles will be taken as 74 + 2 X 
s = 74j inches. Article 74 of the Specifications requires j^ inch ; 
but I ; inch is better practice, since the edges of the web plate are very 
liable to overrun more than -^ 6 inch. Some specifications require 
j inch. 

In the computation of the approximate flange area, the center 
of gravity of each flange will be assumed as one inch from the back 
of the angles. The approximate effective depth is then 74j less 2 
X 1 inch, which equals 72 inches. The approximate stresses in 
the flange areas are : 

, 275 000 X 12 
I* or dead load, - = 4o 000 pounds. 

, , 1 340 000 X 12 
For live load,] - - = 222 000 pounds. 

The approximate flange areas are now obtained by dividing 
these amounts by the allowable unit-stresses for dead and live load, 
which are (see Specifications, p. 8, Article 31): 20 000 and 10 000 
pounds per square inch respectively; and the resulting areas are: 

45 600 

For dead load, ^ = 2 . 28 square inches. 

222 000 
For live load, -T-- = 22. 20 square inches. 

These amounts give a total of 24.48 square inches as the approximate 
net flange area required. 

It will be assumed that one-half the total area, or 24.48 -r- 2 = 
12.24 square inches, is to be taken up by angles. If 12.24 sq. in. 
is distributed- over two angles, then 12.24 -f- 2 = 6.12 square inches 
is the net area for one angle. Of course it is not to be assumed that 
the area of the angle chosen must be exactly 6. 12, but that this 6. 12 
square inches is the approximate area of the angle to be chosen, and 
the net area of the angle (see Specifications, Article 149) must not be 
2 per cent less than this, although it may be greater. 

From Steel Construction, Part I, Table VII, or from the Car- 
negie Handbook, p. 117, a 6 by 6 by f-inch angle gives a gross area of 
8.44 square inches and a net area of 8.44 - 2 X ( + i) X f = 
6 . 94 square inches, J-inch rivets being used and so spaced that two 
rivets are taken out of each angle (see Specifications, Article 63, 
arid Fig. 139). A 6 by 4 by | flinch angle, giving a gross area of 
7.47 and a net area of 6.66 square inches, one rivet-hole being out, 




could have been used, but { jj inch is too thick to punch, and there- 
fore the above angle is chosen. 

The required net area of the cover-plate is now found to be 
24.48 - 2 X 6.94 = 10. 60 square inches. Since the legs of the 
angles arc G inches and the thickness of the web is | inch, the outer 

edges of the angles are 12f inches 
apart; and since the cover-plate 
must extend somewhat over the 
edges of the angle, and the width 
of the cover-plate should be in the 
even inch, the width of the cover- 
plates must be at least 14 inches, 
as shown in Fig. 139. 

On account of the 1-inch rivet- 
holes to be deducted, the real or 
net width of the cover-plate " is : 2 X n + m = 14 2X1 = 12 
inches. The thickness of all the cover-plates at the center is now: 

Fig. 139. Calculation of Size of Angles 
and Cover-Plate. 

;, - 0.885 inch say, 


A thickness of I of an inch is decided upon, for the reason that plates 
are rolled only to the nearest sixteenth of an inch. 

The approximate section at the center has now been determined, 
and is: 

2 Angles by <> by f-inch = 13.88 sq. in. net. 
Cover-plates | inch thick = lO.SOsq. in. net. 

Total = 24.38sq. in. net. 

This approximate section must now be examined, and, if it shows too 
great an excess or a de- ^ 0875 .. |86 

ficiency, must be revised. 
In order to deter- 
mine the effective depth 
the distance between the 
centers of gravity of the 
flanges must first be com- 
puted, the gross areas be- 
ing used. Theoretically, perhaps, the net areas should be used; 
but this is an unnecessary refinement, since the effect on the final 
result is of no practical importance. 

Fig. 140. Determination of Center of Gravity. 



In computing the center of gravity (see Fig. 140), the axis is 
taken at the center of the cover-plates, as this reduces the moment of 
the cover-plates to zero. The distance of the center of gravity of the 
angles from their back (Carnegie Handbook, p. 117, column 6) is 
1 .78 inches'. The distance of this center of gravity from the center 
of the cover-plate, is 1 .78 + 0.875 -^ 2 = 2.22 inches. 

Gross area of the angles = 2 X 8.44 = 1G.88 sq. in. 
" " " " cover-plates = Jx 14 = 12.25sq. in. 

Total = 29.13 sq. in. 

The center of gravity is now found to be 16.88 X 2.22 4- 29.13 = 
1.286 inches from the center of the cover-plate, and 1.286-0.875 
-^ 2 = 0:848 inch from the back of the angle. The effective depth 
h e is 74.25 - 2 X 0.848 = 72.554 inches, and the required flange 
areas are: 

275 000 X 12 

72 . 554 X 20 000 

1 340 OOP X 12 

"727554 X 10 000 

= 2.272 sq. in. for dead load. 
= 22.200 sq. in. for live load. 

Total = 24.472 sq. in. 

The values of the moments, as taken from the curves, must be mul- 
tiplied by 12 in order to reduce them to pound-inches. 

A total of 24.86 square inches is given by the section approxi- 
mately designed, and the difference between that and the section as 
above determined is : (24 . 472 - 24 . 38) -=- 24 . 472 = . 38 per cent, 
and as this is less than 2- per cent (see Specifications, Article 149), it 
may be used without any further change. If there should have been 
a deficiency or an excess greater than 24 per cent, then it would have 
been necessary to revise. In case a revision of section is necessary, the 
size and thickness of the angles generally remain the same as those 
taken in the approximate design, the thickness of the cover-plates 
being decreased or increased as the case may be. 

The total thickness of the cover-plates, | inch, is too thick to be 
punched. In such cases as this, the section is made up of two or more 
plates whose total thickness is equal to that required. If plates of 
more than one thickness are decided upon, then their thickness 
should decrease from the flange angles outward. For the case in 
hand, one plate inch thick and one plate I inch thick will be decided 
upon. The flange section at the center as finally designed is: 







2 Angles 6 by 

6 by f in. 

13.88 sq. in. 



1 Cover-plate 

14 by f in. 

4.50 " 

5 .25 " 

1 Cover-plate 

14 by \ in. 

6 . 00 " 

7 . 00 ' ' 

Total 24. 38 ' ' 


29.13 " 

The above is the section required at the center of the girder; 
for any other point it will be less, decreasing toward the end, where 
it will be zero. Evidently, then, the cover-plates will not be required 
to extend the entire length. The following analysis will determine 
where they should be stopped. If the load were uniform, the moment 

Fig. 141. Diagram Showing Curve of Required Flange Areas. 

curve would be a parabola. Although under wheel loading the curve 
of moments is not a parabola, yet it is sufficient for practical purposes 
to consider it as such. The curve of flange areas, like that of moments, 
is to be considered a parabola (see Fig. 141). 

Let a, = Net area, in square inches, of the outer cover- plate; 

a 2 = Net area, in square inches, of the next cover- plate; 

3 , etc. = Net areas of the other cover-plates; 

A = Net area of all the cover-plates and the flange angle. 
Then, from the properties of the parabola, 

where L = Length of cover-plate in question; 
I = Length of span, center to center; 
a = Net area of that cover-plate and all above it ; and 
A = Total net area of the flange, I of the gross area of the web not being 
considered in this quantity. 



The lengths of the cover-plates for the section above designed (see Fig. 141) 

L, = 61.75 J ^r = 26.45 feet. 

L, = 61.75 = 40.00 feet. 

One foot is usually added on each end of the cover-plate as theoretically 
determined above. The results are also usually rounded off to the nearest 
half-foot. This is done in order to allow a safe margin because of the fact 
that the curve of flange areas is not a true parabola. The final measurements 
of the cover-plates are: 

14 in. by f in. by 28 ft. 6 in. long. 

14 in. by ^ in. by 42 ft. 6 in. long. 

In most cases the cover-plate next to the angle on the top flange only is 
made to extend the entire length of the girder. Although this is not required 
for flange area, it is done in order to provide additional stiffness to the flange 
angles toward the ends of the span, and to prevent the action of the elements 
from deteriorating the angles and the web by attacking the joint at the top 
(see broken lines, Fig. 141, for length of first cover-plate extended). 


1. The dead-load moment equals 469 000 pound-inches; and the live- 
load moment, 4 522 000 pound-inches. Design a flange section entirely 
of angles, if the distance back to back of angles is 45} inches. 

2. The dead-load moment is 3 340 000 pound-inches, and the* live- 
load moment, 21 235 000 pound-inches. Design a flange section using G by 
6-inch angles and three 14-inch cover-plates, the distance back to back of 
flange angles being 78} inches. 

3. In each of the above cases, design the flange section considering 
that i of the web area is taken as effective flange area. (For demonstra- 
tion of the methods to be employed in the solution of this problem, see the 
succeeding text.) 

While the section of a plate-girder is composite that is, it con- 
sists of certain shapes joined together, and is not one solid piece 
nevertheless these shapes are joined so securely that the section may 
be considered as a solid one and its moment of resistance computed 
accordingly. Let Fig. 142 be considered. 

The moment of resistance of the section is: 


in the derivation of which the moment of inertia of the flange about 
its own neutral axis is considered as zero, and A equals the net area of 
one flange. Now, as the values of h e and h seldom differ by more 




than one inch, for all practical purposes they may be considered* as 
equal. The above expression then reduces to : 

M = S x h ( A + -^ ) 

= S X h (net area of flange + one-sixth gross area of web) 

Since the rivet-holes decrease the moment of resistance of the 
web, one-sixth of the gross area cannot be considered, as is theoreti- 
cally indicated in the above formu- 
la. It is common practice to take 
one-eighth, instead of one-sixth, of 
the gross web area. Substituting 
this value in the above equation, 
and transposing, there results: 

Area of flange -I- I gross web area = ^ 

The flange section will now be 
designed for the moments previously 
given, considering of the gross web 
area as efficient in withstanding the 

The gross area of the web is 
74 x f = 27.75 square inches; and 
^ of this is 3.47 square-inches. The 
total approximate amount of flange 
area required is, as in the first case, 
24.48 square inches. 

According to the above formula, 
| of the web area, or 3.47 square 
inches, may be considered as flange 
area, and therefore 24.48 - 3.47 - 
21.01 square inches, is the approxi- 
mate area of the angles and cover-plates of the flange. The ap- 
proximate area of one angle is then 21 .01 -=- (2 X 2) = 5.25 square 
inches. A 6 by 6 by r Vinch angle gives the gross area of 6.43 square 
inches and, two rivet-holes being deducted, a net area of 5.305 
square inches (see "Steel Construction," Part I, Table VIII, or 
Carnegie Handbook, p. 117). As this is quite close to the approxi- 
mate area determined above, this angle will be taken. The ap- 

Fig. 142. Section of Plate-Girder. 




proximate area of the cover-plates is 21.01 - 2 X 5.305 = 10.40 
square inches. As before, the gross width of the cover-plate will 

10 40 
---- ' - 

= . <S67 inch say f inch. 

be taken as 14 inches. The thickness 


The gross area of the angles being 12.86 square inches, and that 
of the cover-plates 12.25 square inches, the center of gravity of the 
section is found, by a method similar to that previously employed, 
to be 1 .10 inches from the center of the cover-plate, or 1.10 0.438 
= . 662 inch from the back of the flange angles. This makes the 
effective depth 72.93 inches. 

For this section, the true live-load flange stress is (1 340 000 X 
12) -T- 72 . 93 = 221 000 pounds, and the actual dead-load flange stress 
is (275000 X 12) -1-72.93 = 45400 pounds. The actual areas 
required for the live and dead load are 22. 10 and 2.27 square inches, 
which are obtained by dividing the above flange stresses by 10 000 
and 20 000 pounds, respectively. The total required area is the sum 
of the two areas above, and is equal to 24.37 square inches. The 
total area required in the flange angles and cover-plates is therefore 
24.37, less the gross area of the web, 3.47, which leaves 20.90 
square inches. The same angles as decided upon before will be used. 
This gives a required area for the cover-plates of 20.90 10.61 = 
10.29 square inches. The required thickness is then 10.29 + (14 
2) = 0.857 say f inch. The following section of the flange will 
therefore be decided upon: 




2 Angles 6 in. by 6 in. by T 9 ^ in. 
1 Cover-plate 14 in. by H in. 
1 Cover-plate 14 in. by \ in. 

10. Gl sq. in. 
4 . 50 " 
6 . 00 " 

12.86 sq. in. 
5.25 " 
7.00 " 

Total = 

21.11 " 

25.11 " 

As the total net area above is within 2^ per cent of the required net 
area, that section will be taken (see Specifications, Article 149). Note 
that in this case, the thickness of the cover-plates in the final design 
is the same as that determined in the preliminary design. 
Also note that the total net area is about 4 square inches, or 20 per 



cent, less than in the flange as first designed, in which case none of the 
area of the web was considered as withstanding the bending moment. 
The ^-inch cover-plate on the top flange will extend the entire 
length of the grider, and is therefore 62 feet 9 inches long. The 
lengths of the other cover-plates are: 

For i-inch plate at the bottom, L = 01 .75 -y' ^ ' '^ = 43 . 5 feet. 

For each f-inch plate, L = 61.75 ^^lT = 28 - 5 feet - 

One foot should be added to each of the above lengths at each end, 
thus making the total lengths 45 feet (i inches and 30 feet 6 inches, 


1. If the span is 63 feet center to center, compute the length of the 
cover-plate. The section consists of two angles 6 by 6 by f in.; one cover- 
plate 14 by \ in.; and one cover-plate 14 by % in.; two rivet-holes being taken 
out of each angle and each cover-plate. 

2. If the span is 87 ft. 9 in. Center to center, compute the length of the 
cover-plate if the flange consists of two angles 6 by 6 by | in., and four cover- 
plates 16 by j'g- in., two rivet-holes being taken out of each angle and each 

In determining the area of plates, the tables in the Carnegie 
Handbook, pp. 245 to 250, are convenient. In order to obtain plates 

F F F F C F 

Pig. 143. Diagram Illustrating Transference of Shear from Web to Flanges by Rivets. 

whose widths are greater than 12f inches, see the note in the right- 
hand column on page 250, Carnegie Handbook. For another pres- 
entation of the above subject-matter, see "Steel Construction," Part 
IV. pp. 252, 254, and 261. 

The spacing of the rivets in the flanges is a matter of considerable 
importance; the shear is transferred from the web to the flanges, 
where it becomes flange stress. This is done by the rivets, each rivet 
taking as much flange stress as is allowed by the Specifications. The 
conditions are similar to those shown in Fig. 143, where V represents 




an object exerting a pull on a long, thin plate A - A which has, at 
various points along this length, small objects F attached to it by 
means of pegs r - r. These small objects F hold the plate A- A in 
equilibrium. Here V represents the shear which tends to cause the 
movement; A- A, the web; r-r, the rivets; and 2F the amount of 
flange stress taken by each rivet. 

At section c - c the total amount in the web to be trans- 
formed is 2F; at section b - b 
it is 10F. From this it is seen 
that enough rivets r-r must be 
put in between the sections b - 
b and c - c to take up 10F 
2F = 8F; hence it is proved 
that the rivets between any two 
sections of the flange take up 
the difference in flange stress 
between those two sections. 

The discussion just given 
will be the means of giving us the 
number of rivets required between any two sections; but it does not 
give us the rivet spacing between these two sections. In order to 
determine the rivet spacing at any particular point, the following 
analysis is presented (see Fig. 144). 

Let A/\ = Moment at one section, 

M 2 = Moment at another section nearer center of girder than the section 

where Af, occurs; 

V = Shear at section where M , occurs; 
v = Amount of flange stress one rivet can transfer; or it is the stress on 

one rivet; 

s = Distance between the two sections; 
n = Number of rivets between the two sections. 


Fig. 144. Determination of Rivet Spacing. 

- Flange stress due to moment M,; 


_ 2 = Flange stress due to moment M.,; 

2 i = Difference of flange stress between the two sections; 

( 2 _ J \ -j- v = n, Number of rivets required in space ...(!) 




If the above sections be taken close enough together so that 
the number of rivets required 1 (that is, n = 1), then V 
can be considered as constant between the two sections, and then 
the moment M 2 = M v + Vs (see Article 44, Part I). Substituting 
in Equation 1, above, there results: 

/M. + V* M, \ 

from which, 

rh e 

which is the formula for the rivet spacing in the vertical parts of the 
flanges of any girder, providing the flange is not subjected to localized 
loading. It is to be used for the rivet 
spacing in both the top and bottom 
flanges of through girders, but not in 
the top flanges of deck plate-girders for 
railroad service. It is to be used, how- 
ever, in the bottom flanges of deck plate- 
girders for railroad service. The dis- 
tance A e is not ordinarily used, the 
distance between rivet lines being used 
instead (see Fig. 145). The rivet spacing 

Fig. 145. Determination of Rivet j n the cover .pl ates an d horizontal legs of 

the angles is made to stagger with that 

in the vertical legs, and usually the staggering is with every other rivet 
in the vertical flange. The term stagger signifies that the rivets in 
the top flange are not placed opposite the rivets in the vertical legs 
of the flange angles or, that in case there are two lines of rivets 
in the vertical legs of the angle, a rivet near the outer edge of the 
cover-plate is placed in the same section where a rivet occurs near the 
lower edge of the vertical legs of the angle, and vice versa. 


1. Determine the rivet spacing at a section where the shear is 147 200 
pounds, the value of one rivet 4 920 pounds, and the effective depth of the 
section 84} inches. 

ANS. 2.82 inches. 

2. Determine the stress on a rivet at a section where the shear is 
299 400 pounds, the spacing 2^ inches, and the effective depth of the girder 
84} inches. 

ANS. 8 870 pounds. 




The rivet spacing is usually determined at the tenth-points; 
and a curve is plotted with the spacing as ordinates and the tenth- 
points as abscissse. The rivet spacing at any intermediate point can 
be determined from this curve. When one-eighth the gross section 
of the web is considered as flange area, then only that proportion of 
the shear which is transferred to flanges is to be considered in com- 
puting the rivet spacing, on account of the fact that some of the 
shear is transferred directly to the bending moment in the web. 

In order to determine the distance between rivet lines, the 
gauge, or distance out from the back of the angles to the place where 
the rivets must be placed, must be known for different lengths of legs. 
Table XXI gives the standard gauges, and also the diameter of the 
largest rivet or bolt which is allowed to be used in any sized leg. No 
gauges should be punched otherwise unless your large experience or 
instructions from one higher in authority demand it, and this should 
be so seldom that indeed it might be said never to be necessary. 


Standard Gauges for Angles 
(All dimensions given in inches) 


MAXI- || 




oi{ BOLT 


































































*When thickness is J- inch or over. 




Fig. 146. Determination of 
Distance between 

The distance between rivet lines for the girder being designed 
(see Fig. 146), is, in the first case: 

= 74.25 - (2 X 2i + 2J) 
= 67.00 inches. 

In the second case, where | of the web is 
considered, the above distance is 74.25 
(2 X 2J + 24) = 67.25 inches. The compu- 
tations and the rivet spaces at the tenth- 
point, and at the ends of the cover-plates in 
the bottom chord of the plate-girder, are 
shown for each case in Table XXII. The 
value of v is the value of a |-inch rivet in 
bearing in a f-inch web (see Specifications, 
Article 40, and Carnegie Handbook, p. 195, 
second table). This value is 4 920 pounds. 

In the first column, 7.98 indicates that 
the end of the cover-plate next to the flange 

is 7.98 feet from the end of the girder, and that this section is taken 
just to one side of that point, the side being that nearest the end of 
the girder. In a similar manner, 7.98+ indicates that the section 
'is taken to that side of the point which is nearest the center of the 
girder. A like interpretation should be placed on 15.55 and 
15.55 + , the point under consideration in this -case being the end of 
the outer or top cover-plate. 

In the fifth column, values are given which indicate that portion 
of the shear which is transferred to the flanges. For example, 
Q7 700 

40 " 10 en = 74 700 ' and the difference between 97 70 and 

74700 represents that portion of the shear which is taken up 
directly by the web in the form of bending moment. An inspection 
of the headings of the third and fourth columns will tend to make 
this matter clearer. 

Where there is local loading, as in the top flange, the rivets, in 
addition to the stress caused by the transferring of web stresses, are 
stressed by the vertical action of the angles being pressed downward 
by the ties and the consequent upward pressure of the web. Accord- 




Rivet Spacing in Bottom Flange 

Flange Taking All the Moment 






















61 300 67 




44 200 67 



I See Specifica- 


28 600 - 




\ tions, Art. 54 

One-Eighth of Web Area Considered 
hi ~ 67.25 inches; v = 4 920 pounds 







(Sq. Inches) 

(Sq. Inches) 









97 700 



74 700 



90 000 



67 900 


7.98 + 
















56 000 


15.55 + 




58 100 



61 300 


21.11 . 

52 700 



44 200 









24 600 


ing to Article 15 of the Specifications, the weight of one driver is 
distributed over three ties (see Fig. 147). 

Let -j- , = w, the load per linear inch caused by one wheel W, which 

load is assumed to be uniformly distributed over the dis- 
tance 7; 

ws = t'i, the vertical load or stress that comes on one rivet in the 
space s; 

v = -T , the stress on a rivet due to the distribution of flange 

stresses when s is a space, and V the shear at that point. 

When these two stresses act on the rivet, the maximum stress will 
be v u , the ultimate amount that it is allowed to carry, and this will 
act as shown in Fig. 147 : Then, 




r- ) + (w*) 2 

from which, 

which gives the spacing at any point in the girder flange under 
localized loading. Note that if w equals zero that is, if there is no 
localized loading there results: 

which is the same as previously deduced for flanges with- 
out localized load- 

The rivet spacing 
for the top flange 
of the girder which 
is being designed 
is given in Table 
XXIII. Here W 
= 20000; / = (3X 
7 + 3 X 6) = 39 

. , 20 000 

inches; w = ~ 

= 513; h r = 67 
inches; and v w = 
4 920 pounds, which 
is the bearing of a 
f-inch rivet in the 
1-inch web. The 
top cover-plate is 
run theen tire length 
of the span. 

Fig 147. Rivet Spacing Determined by Stresses Distributed 
under Localized Loading. 




Rivet Spacing in Top Flange 

Flange Taking All the Moment 







262 GOO 


3 080 000 

1 825 



' ' 


2 140 000 

1 550 



79 300 

1 390 000 

1 285 

3 . 83 



61 300 

840 000 




" . 


435 000 





181 500 



One-Eighth of Web Area Considered 
= 513; h t = 67i inches; v u = 4 920 pounds. 







262 600 


2 100 000 

1 538 



262 600 


1 450 000 




262 600 


985 000 

1 100 



262 600 


695 000 



15.55 + 

262 600 

58 100 

765 000 

1 014 



262 600 

52 700 

616 000 




262 600 

38 000 

320 000 




262 600 


134 000 



The points other than the tenth-points referred to in the first column 
are for sections taken just to the left and right of the top cover-plate. 
The values of the reduced shears given in the third column are ob- 
tained as has been previously explained. Although the rivet spacing 
in the lower flange is considerably greater than that in the upper 
flange, and accordingly a smaller number of rivets would be required, 
yet the spacing in the lower flange is made the same as that in the 
upper. Convenience in the preparing of plans and facility in manu- 
facture make this action economical. Theoretical spacing greater 
than 6 inches should be dealt with according to Article 54 of the 

The values of the rivet spacing given in Tables XXII and XXIII 
are plotted in Fig. 148. Note that the effect of the localized loading 
is to decrease the rivet spacing, and also note that the effect increases 
from the ends toward the center. 






"d 8 


Note.-5ccond Cover Plate oi 

Top Flange Extends Lntire 
Length, of Girder. 

o \ e 3 ^ 5 

Fig. 148. Plotted Values of Rivet Spacing Given in Tables XXII and XXIII. 




The size of the flange angles and the width of the cover-plates 
for different spans, are a matter of choice. Once the size is deter- 
mined, the thickness can be computed. The sizes very generally 
adopted in practice are as follows: 



15 to 20 feet 

5x3^ inches 





' ' 



6x6 ' 

' ' 



6x6 ' 





6x6 ' 



100' 120 



For another method of the presentation of this subject, see 
"Steel Construction," Part IV, pp. 264 to 268. 


1. Determine the rivet spacing for the top chord of a plate-girder, 
loading E 40, and 7 by 9-inch ties being used. The web is f inch thick; 
distance from back to back of angles, 6 feet 6^ inches; flange angles, 6 by 6 
by ^-inch; and cover- plate, 14 by jj-inch, two J-inch rivets'being out of each. 
First, consider the flange as taking all the bending moment; and second, 
consider one-eighth the gross area of the web. The total unreduced shear 
is 80 000 pounds in both cases. 

ANS. 3 . 21 inches; 3 . 76 inches. 

74. Lateral Systems and Cross=Frames. There are two methods 
in use in common practice in determining where the panels of the 
lateral bracing shall fall namely, (1) To choose the number of panels 
so that the panel points come opposite the stiffness, and (2) to choose 
the number of panels so that the placing of the panel points is inde- 
pendent of the stiffener spacing. The lateral systems should be of 
the Warren type; and in both of the above cases the angles that the 
diagonals make with the girder should not be greater than 45 degrees. 
Also, it is best to have all panels the same length and to have an equal 
number of panels. This latter condition will simplify the drafting 
very much, since one-half of one girder can be drawn and the other 
half will be symmetrical, the opposite girder being similar to the one 
drawn, but being left-handed. 

The members of the lateral systems will take tension or com- 
pression according to the direction the wind blows. Cross-frames 
are placed at intermediate points to stiffen the girders. These are 



diagonal bracings (see Plate II), and are placed at certain intervals 
according to the judgment of the engineer. Good practice demands 
that their number should be about as indicated below : 



to 20 feet 

2 . 

20 to 35 " 


35 to 70 " 


70 to 85 " 


85 to 110 " 


The above is not intended to serve as a hard and fixed rule. Varia- 
tions from the limits given are to be made as the case demands. In 
all cases they are put at the panel points of the bracing, the top and 
bottom parts acting as sub-verticals in the lateral system. Also, the 
cross-frames should divide the span into equal parts if possible. In 
cases where that is not possible, the shortest divisions should come 
near the ends of the spans. 

If the panel points are to be located at the stiffene'rs, the number 
of panels is a function of the depth of the girder (see Specifications, 
Articles 47 and 48) . In this case the number of panels is given by : 

,. _ Span in inches. 

Depth of girder in inches' 

no fraction being considered. As an example, let it be required to 
determine the number of panels in a girder 85 feet center to center 
of bearings, the depth being 90| inches back to back of angles. Then, 

85 X 12 
N = qn~~9~~ = 11-3, or, say, 11 panels. 

Each panel will then be 92.8 inches long. This, according to Article 
47 of the Specifications, being greater than 5 feet, would not be allowed 
as a space between two stiffeners; but one stiffener can be placed in 
between, and then the panel points will come at every other stiffener. 
The cross-frames should be five in number. 

The arrangement of panels and cross-frames is shown in Fig. 
149. Here the cross-frames are marked C. F., and the broken lines 
represent the low r er lateral system. 

In case an even number of panels were desired, then ten would 
be the number chosen and the general arrangement would be as 
shown in Fig. 150. The length of a panel would be 85 X 12 4- 




10 = 102 inches, or 8 feet 6 inches, which would allow of one stiffener 
in between and still keep the stiffener spacing within the limit of 
5 feet. 

The cross-frames at the ends of the span are designated as end 
cross-frames, and those in between are designated as intermediate 

In case the spacing of the stiffeners is not required to be such as 
to coincide with the panel points of the lateral bracing, the panel 
length will depend upon the spacing of the girders, being equal to or 

Fig. 150. 
Arrangements of Panels and Cross- Frames. 

greater than the spacing in order to keep the angle which the diagonals 
make with the girder less than 45 degrees. In this case, 

Span in feet 


Distance center to center of girders in feet 

For the girder considered on page 1 74, the number of panels would be 

11.3 or, say, 11 panels if odd numbers were to be used, 

and 12 if even numbers were to be desired. 

For the case in hand, the panel points of the bracing will be taken 

at the stiffeners, and an even number of panels will be used. Then, 

61.75 X 12 


9. 98 (say 10). 

The arrangement of the panels and cross-frames, and also the maxi- 
mum stresses in the diagonals, are shown in Plate II, the stresses being 
determined according to Article 50, Part I, "Bridge Analysis," and 
Article 24 of the Specifications. All the wind is taken as acting on 
one side of the bridge; and no overturning effect, either on the girder 



- 175 

or on the train, is considered. Also, note that the wind stresses in the 
flanges a-re not considered. Should the student determine these, he 
will find them too small to be considered according to Specifications, 
Article 39. 

Before designing the lateral diagonals which consist of one or 
two angles, Articles 31, 33, 34, 35 (last portion), 38, 40, 63, and 83 of 
the Specifications should be care- 
fully studied. The upper lateral ^ e^ft. j 

bracing is to be designed first. 
Carnegie Handbook, pp. 109 to 
119, is to be used. 

The member U U 1 must be 
designed for a compressive stress 
equal to 23.20 + 0.8 X 20.6 
= 39.68, and a tensile stress of 
20.6+0.8X20.6 = +37.08. The 
length of the diagonal measured 
from center to center of girders 
is 1/6.5' +6.2 2 - 9 feet, or 108 
inches. In reality the length is not 

108 inches, as the cover-plate takes off a certain amount, as shown 
in Fig. 151. The true length, which is to be taken as a column length 
in designing, is 108 2y, and y is readily computed to be 9.70 inches, 
thus making the true length 88.6 inches. The least allowable rectan- 
gular radius of gyration is obtained from the relation that the greatest 

Fig. 151. Determination of Length of 
Diagonal in Lateral Bracing. 

value of - = 120, and therefore the least value of r = 



r 120 

It will be assumed that a 6 by 4 by r 9 6 -inch angle with an area of 
5.31 square inches will be sufficient. Here the length equals 88 . 60 
inches, and the least rectangular radius of gyration is 1 . 14; hence, 

oo *r\ 

P = 13 000 - 60 X ~~ 
1 .14 

= 8 330 pounds per square inch, 
on QAQ 

The required area is = 4 . 73 square inches. As the angle 
8 ooO 

assumed has an area of 5.31 square inches, which is considerably 
greater than the 4.73 square inches required, this angle cannot be 
used, and other assumptions must be made until the area of the angle 




assumed and the required area as computed are equal or very nearly so. 

A 6 by 4 by ^-inch angle with an area of 4.75 square inches will 

now be assumed. The length is 88 .60 inches as before, and the least 

rectangular radius of gyration is 1.15. The unit-1'oad P = 8340 

39 360 
pounds per square inch, and the required area is ' * = 4 . 72 square 

o o4U 

inches. As the area of the angle assumed and the required area as 
computed are very close, this sized angle will be used. 

The section must now be examined for tension, and in order that 
both legs of the angle may be considered as effective section, both legs 

must be connected at the end. 

, i -11 u 36 700 
1 he area required will be 

18 000 

2.04 square inches. 

o o \o / \ 
[_ o_ _P\gx_ -X \ 

u i 1 

Cover Plate -j ( 

Considering one rivet-hole is taken out of the 
angle, the net area is 4 . 75 1 X 
(1 + s ) X $ = 4 . 25 square inches, 
which is amply sufficient. 

If the 4-inch leg only were as- 
sumed to be connected, the gross 
area would be 4 X $ = 2. CO 
square inches, and the net area 
would then be 2.00 - 1 X (A + 
s) X ^-inch = 1 .50 square inches, 
which is not sufficient. If the 
6-inch leg were connected, the 
area would be sufficient. See Fig. 
152 for method of connection and 

The number of rivets required (see Specifications, Article 38 
and 40) is computed as follows: If the member were not subjected 
to both tension and compression, the number of rivets required in 
single shear would be: 


(9 000 + 50 per cent of 9 000) X . 6013 
_ 39360 

= 4. 86 (say 5). 

But according to Article 38 of the Specifications, this number must be 
increased 50 per cent, and accordingly 4.86 X U =7. 29 (say 8) 

Fig. 152. Method of Connecting Angle Legs 
in Lateral Bracing and Cross-Frames. 


JM . 

S 5 

K f 
W j? 


M - 


w 2 
SB 75 
5 -2 

Cd = 
O * 


shop rivets are to be used. In the above formula, 0.6013 is the area 
of the cross-section in square inches of a f -inch rivet. In order that 
both legs should be connected, a clip angle as shown in Fig. 152 is 
used ; and the same number of rivets must go through both legs of the 
clip angle, since the stress in the vertical leg of the main angle is 
transferred to the clip angle and from there into the connecting plate. 

The above number of rivets makes the joint safe so that it will 
not shear off in the plane between the connection plate and the hori- 
zontal leg of the angle. The joint must also be designed so that 
there will be sufficient rivets in bearing to prevent them from tearing 
out of the connecting plate. The number required is: 

= 39 360 _____ 

~ (15 000 + 50 per cent of 15 000) X | X f 

- I 39 3GO 
= ~7ll80 

= 5.34. 

This 5.34 must be increased 50 per cent, making a total of 5.34 
X 1.5 = 8.01 = say, 8 shop rivets as before. 

The above rivets are shop rivets, since it is assumed that the span, 
being a small one, will be riveted complete in the shop and shipped to 
the bridge site ready to place in position without any further riveting. 
In case the girders are shipped separately, then the lateral bracing 
must be riveted up in the field ; and according to last part of Article 
-40 of the Specifications, the rivets, being field rivets, must have the 
allowed unit-stresses reduced one-third, which is equivalent to having 
the number of shop rivets increased 50 per cent. This will make the 
required number of field rivets 8 X 1.5 = 12. 

As a rule, the connection plates are f inch thick, seldom more. 
Also, the members of the upper lateral system are connected on the 
lower side of the connection plate in order not to interfere with the 
ties. Note that the use of the clip angles requires a smaller connec- 
tion plate than would be necessary if these angles were not used, since 
in the latter case all the rivets must then be placed in one row in the 
horizontal leg of the angle. 

The number of rivets required in the connection plate and the 
flange of the girder must be sufficient to take up the component of 
that member parallel to the girder. For the case in hand, the num- 
ber (see Fig. 153) is: 




x _6 1 2 
T ~{TO' 
from which, 

x = 5.5 (say 6) rivets. 

Additional rivets should also be put in, in order to take up the compo- 
nent of the other lateral diagonal which meets at this point. 

The member L T / U l is to be designed for a maximum compressive 
stress of 20 . 6 + . 8 X 16 . - - 33 . 4. A 6 by 4 by -, Vinch angle 

with an area of 4.18 square 
inches will be assumed. The 
least rectangular radius of gyra- 
tion is 1.1G. The unit allowable 
load is: 

P = 13 000 - 60 X f^rt = 8 420 
1 . lo 

pounds per square inch. 

The required area is 

8 420 

3.97 square inches. As this is 
very near the area assumed, and 
as trials with other angles do 
not give required areas which 
come any closer, this angle will 
be used. 

The rivets required in single 
shear are : 


6^ ft. 

55 nv. 

Fig. 153. Calculation of Rivets in Connection 
Plate and Flange of Girder. 

33 400 

X 1.5 = 6.21 (say 7) shop rivets, and 

6.21 X 1.5 = 9.3 (say 10) field rivets. 

The rivets required in bearing in the f-inch connection plate are: 

jj-ngh X 1.5 = 6 . 78 (say 7) shop rivets, and 
6.78 X 1.5 = 10.17 (say 11) field rivets. 

The above computations show the joint to be weakest in bearing, 
and therefore 7 shop or 11 field rivets must be used. It is not neces- 
sary to investigate this member for tension, as the computations for 
the first diagonal indicate that the area will be sufficient, both legs 
being connected. 

The member UJJ 2 ' must be designed for a maximum compres- 
sive stress of 16.0 + 0.8 X 14.1 = -27.28. A 6 by 4 by f-inch 



angle with an area of 3.61 square inches and a least rectangular 
radius of gyration of 1.17 will be assumed. The unit-stress P, as 
computed from the formula in the Specifications, is 8 400 pounds per 

27 280 

square inch; and the required area is -- ~ = 3 . 23 square inches. 

8 460 

This angle will be used, as the given and required areas are close 
together, and as the next smaller angle a 6 by 3i- by -inch angle 
with an area of 3.42 square inches gives a required area of 3.58 
square inches, thus being too small. 

The rivets required in single shear are : 

27 280 

X 1.5 = 5 . 04 (say 5) shop rivets, and 
o 1UU 

5.04 X 1.5 = 7. G (say 8) field rivets. 
The rivets required in bearing in a f-inch web are : 

27 280 

X 1.5 = 5 . 54 (say 6) shop rivets, and 
7 ooO 

5.54 X 1.5 = 8.3 (say 9) field rivets. 

In order to make the joints safe, 6 shop or 9 field rivets should be used. 
The member U 2 'U 2 must be designed for a maximum compressive 
stress of 9.6 + 0.8 X 8.0 - -16.00. A 3^ by 3 by f-inch angle 
with an area of 2.30 square inches and a least rectangular radius of 
gyration of 0.90 will be assumed. The unit-stress P is 7 090 pounds 

. 16000 
per square inch, and the required area is _ ft( ~- = 2 . 26 square inches. 

As the required and the 'actual areas are very close together, this 
angle will be used. 

The rivets required in single shear are : 

X 1.5 = 2.96 (say 3) shop rivets, and 
o lOU 

2.96 X H = 4.44 (say 5) field rivets. 

By computation similar to the above, it is found that 4 shop or 5 
field rivets are required in bearing. Since the bearing requires the 
most rivets to make the joint safe, 4 shop or 5 field rivets must be used. 

If the Specifications would have allowed a 3! by 3^ by j 5 <pinch 
angle with an area of 2.09 square inches, this angle would have 
exactly fulfilled the requirements, the required area being 2.09 
square inches. 

The member U 2 U a r must be designed for a maximum compres- 



sive stress of 8.0 + 0. 8X4. 1= -11.28. A 3 by 3 by 1-inch 
angle with an area of 2.11 square inches and a least radius of 
gyration of 0.91 will be assumed. In this case the unit-stress is 
7 160, and the area required is 1.58 square inches. The required 
area is considerably less than the area of the angle assumed; but it 
must be used, since it is the smallest allowed by the Specifications, 
which require that the material shall not be less than f-inch, and 
from Table XXI it is seen that 3 inches is the smallest size leg in which 
a | -inch rivet can be used. 

The stresses in all the members of the lower lateral system are 
less than the stresses in the member just designed, and therefore all 
members of the lower lateral system will be made of one 3 by 3 by 
f-inch angle. 

For the last member designed in the upper lateral system, and 
for all members in the lower lateral system, 3 shop or 5 field rivets 
will be required at the ends. These are more than sufficient to take 
up the stress, but it has been found that less- than three rivets do not 
make a good joint. 

The stress sheet, Plate II, shows the general arrangement of the 
lateral system, the number of rivets in the connections and also in the 
connection plates where they join the flanges. 

The intermediate cross-frames do not lend themselves to a theo- 
retical design, since the stresses which come upon them are not easily 
ascertained. It is good practice to require that all members be of 
the sizes as given below: 


(in Feet) 

(in Inches) 


30 to 65 
65 to 110 

3i-x 3* x f 
4 x 4 x f 


4 5 

The angles in the intermediate cross-frames will therefore be 3\ by 
3\ by -inch. 

The end cross-frames (see Fig. 154) act in a manner somewhat 
similar to the portal bracing in a bridge, since they transfer all the 
wind which comes on the top chord and on the train to the abutment. 
This load, which acts at the level of the ties, is in this case (see Article 
24 of the Specifications) : 




P = 

600 X 61 ft. 9 in. 

18 525 pounds. 

It is usually assumed that half of this is' transferred to the point a by 
means of a-b, and from there down a-b' to the masonry. The other 
half goes directly down b-a r to the masonry. This causes stresses as 
shown in Fig. 154. Note that the stress in a-b will always be com- 
pression; but the stresses in the diagonal will be either tension or 
compression according to the direction the wind blows. The mem- 
ber a-b will be a 
3^ by 3o by f -inch 
angle. To form 
the connections at 
its end, 3 shop or 
5 field rivets will 
be used. 

The maxi- 
mum compressive 
stress for which 
the diagonals are 
to be designed is 
12.70 + 0.8X12.70 
= -22.86. Here 
the length is 108 

inches if the angle tends to bend one way; but if it bends as 
shown by the broken lines in Fig. 154, the length will be one-half 
of this. For this reason, angles with unequal legs should be used, 
the longer leg extending outward. This allows the greatest rectan- 
gular radius of gyration to be used. 

A 4 by 3 by 

and a radius of gyration of 1 .25 will be assumed. The unit-load P 
is computed to be 8 750 pounds, and the required area is therefore 
22 860 -r- 8 750 = 2.61 square inches. This does not coincide very 
closely with the given area, but will be used since this angle comes 
nearer to fulfilling the condition than any of the other sizes rolled. 
The joint will require more rivets in bearing than in single shear. 
It is not necessary to perform the complete computations in order to 
determine this, since a comparison of the values of a rivet in single 
shear and in bearing shows that the value in bearing is less than that 

Fig. 154. Action of End Cross-Frames. 

1 6 -inch angle with an area of 2.87 square inches 




in single shear, and therefore the number of rivets required in bearing 
will be greater than that number required in single shear. The 
number of rivets required in bearing is: 

22 860 
4920-XTB =^ 4 ^p rivets, and 

4.00 X 1.5 = 6 field rivets. 

75. The Stiffeners. According to Article 47 of the Specifica- 
tions, these should be placed at certain intervals whenever the unit- 
shear is greater than 

5 = 10 000 - 75 X ~ = 10 000 - 14 800 = -4 800 pounds. 

This negative 
sign signifies that 
whenever the 
unit shearing 
stress is greater 
than zero, the 
stiffeners must be 
placed through- 
out the entire 
length of the 
span at distances 
not to exceed 5 

The interme- 
diate stiffeners 
should have the 
outstanding leg 
long enough to 
give good sup- 
port to the flange angle (see Fig. 155). The filler bars or fillers 
are put in so as to allow the stiffener angles to remain straight 
throughout their entire length; otherwise they will have to be 
bent as shown in Fig. 156. This bending is called crimping. 
Stiffeners must also conform to Article 48 of the Specifications. 
This would require a different sized stiffener at each point, and also 
a different number of rivets in each stiffener. This is not usually 
done in practice. In practice the stiffener for the first intermediate 

Fig. 155. Use of Straight Stiffeners, with Filler Bars. 




point is designed, and the remainder are made the same in size and 
have the same number of rivets. An exception to this is where a 
stiffener comes at a web splice. In this case the size is usually kept 
the same, but the number of rivets is changed somewhat to conform 
to the requirements of the splice design. 

The second intermediate stiffener comes at the first tenth-point, 
and is 6. 175 (say 6.2) feet from the end, since it is at the first panel 
point, or opposite the first panel point, of the lateral system. The 
first stiffener will be 3 . 1 feet from the end ; and scaling off the value 
of the shear at this point (see Fig. 134), it is found to be 108000 

Fig. 156. Crimping of Stiffener Angles 
'where No Filler Bars are Used. 

Fig. 157. Section of Intermediate Stiffener 

pounds. Here the length I to be used in the formula for the unit 
allowable compressive stress is 74J 2 X f = 72.75 inches, the f 
being the thickness of the flange angle. The section of the 
material which according to Article 48 of the Specifications is to be 
considered as a column, is shown in Fig. 157. The assumed 
column cannot bend about the axis B -B, but about the axis A- A, 
and therefore the radius of gyration about the axis A - A must be 
computed. The moment of inertia of the fillers and the web plates 
about their own axes is considered as zero. 

A 4 by 4 by ^-inch angle with an area of 3 . 75 square inches will 
be assumed to be sufficient to withstand the stress. The moment 
of inertia of this and the filler bars and the web portion is 

/A_A = 2 (5.55 + 3.75 X 2.12- 4- 3.00 X 0.563 2 ) = 46.70 r 




The radius of gyration, then, is, */ ' -= 1.764, and the unit-stress 

computed with this value and a length of 72 . 75 inches is 8 140 pounds. 
The required area is now determined to be 108 000 -=- 8 140 - 13 . 27 
square inches. The value 15.00 used in the above computation for 
the radius of gyration is the value of the area of the angles, the filler 
bars, and the web portion. A 5 by 3i-inch angle with the 5-inch leg 
out would have given better support to the flange, but would not 

make so good a job, as it would 
have extended about inch be- 
yond the curved part of the hori- 
zontal leg of the flange angle. 

The bearing determines the 
number of rivets in this case. 



\fiic ater 

Less thair 


Fig. 158. Rivets Placed in Two Rows to 
Give Necessary Number and Spacing. 

The number is 108 000 -=- 4 920 
= 22 shop rivets in the web. 

The angle must now be inves- 
tigated in order to see if these 22 
rivets can go in one row without 
being closer together than 2| 
inches, which is three diameters 
of the f-inch rivet. The total 
length in which these rivets must 

be placed is 72.25 inches, and therefore we have 72.25 -r- 22 = 3.3 
inches as a spacing. Since this is greater than 2| inches, 22 
rivets can be placed in one row. If the spacing as determined 
above had been less than 2f inches, it would have been neces- 
sary to use two rows of rivets spaced as shown in Fig. 158; and then 
the distance center to center would be more than 2f inches, although 
the spacing in a vertical line would be less than that. 

The four angles at the ends of the girders are called the end 
stifleners. These are placed in pairs on opposite sides of the web 
(see Plate II, Article 74). 

The total end shear is 117 800 pounds, and this is assumed to be 
carried by the two pairs of end-stiffener angles, each carrying one- 
half. This amount would require lighter angles than the angles 
used for intermediate stiffeners It is the customary practice tc 
make them the same size and thickness as the intermediate stiffeners, 




additional strength being allowed in order to withstand the effects 
of the end cross-frame when in action. 

The bearing determines the number of rivets required in each 

117 800 
pair of stiffeners. The number required is -- = 12 shop 


Some engineers arbitrarily choose the stiffeners regardless of 
the shear, enough rivets, however, being put in the end stiffeners to 
take up all the shear; and the spacing in the intermediate stiffeners is 
made the same. One noted engineering firm determines its stifFeners 
according to the following: 








A T i : 

4 in. 


3 x 3 



3 X 3 


5 in. 

Over | in. 





3* rX3i 

;n x si 



6 in. 

Less than | in. 




5 x3$ 


8 in. 





6 x4 


No rational method has as yet been determined for ascertaining 
the stresses in the stiffeners of plate-girders. Results obtained by 
placing extensomcters on the stiffeners of actual plate-girders appear 
to indicate that the stresses are very small, in fact in most cases not 
being greater than 1 500 or 2 000 pounds per square inch. 


1. Design, according to Cooper's Specifications, the end stiffeners it 
the shear is 150 000 pounds, the distance back to back of angles is 6 feet 6\ 
inches, the web f inch thick, and the flange angle 6 by 6 by Wnch. Use fillers. 

2. Design the intermediate stiffeners for the girder of Problem 1, 
above, where the shear is equal to 75 000 pounds. Use crimped stiffener angles. 
Note that in this case the angles lie close against the web, no filler bars being 
used in between. 

76. The Web Splice. Web splices are required because of the 
fact that wide plates cannot be rolled sufficiently long. Web splices 
should be as few as possible, and good practice demands that they be 
placed at the same points as the stiffener angle. 

The tables on page 30 of the Carnegie Handbook give the extreme 
length of plates which can be procured for any given width. The 




length of plates for widths which are not given in these tables, should be 
taken equal to the length of the next plate given whose width is less 
than that of the desired plate. From the first table it is seen that a 
74 by f-mch plate can be rolled up to 400 inches, or 33 feet 4 inches, 
in length. Therefore, if the girder under consideration is spliced 

at the center, the web plates will be required to be- ^ = 30 

feet 10?r inches, which value does not exceed the 33 feet 4 inches as 

given above. 

According to Articles 46 and 71 of the Specifications, a plate must 

be placed on each 
side of the web as 
shown in Fig. 159, 
and enough rivets 
placed in each side 
to take the total 
shear. The total 
thickness of both 
plates, and also 
their length, must 
be sufficient to 
stand the total 
shear, but must 
not be less than 
f inch. 

The total shear 

Section A- A 
Fig. 159. Splice Plates Placed on Each Side of Web. 

at the center of 
the girder under consideration (see Fig. 134, p. 150) is 28 600 pounds. 

The area required 

, . -, ,. . . . 28600 
each 01 the two splice plates is : - 

= 1 . 59 square inches ; and as their length is 62 . 25 inches, the 
thickness must be 1 .59 H- 62.25 = 0.0255 inch, but they must be 
made f inch thick according to the Specifications. The width should 
be somewhat greater than twice the width of the stift'ener angle leg. 
This would make the width in this case about 10 inches. 

The bearing governs the numl>er of rivets required in this case, 
and they are 28 600 -f- 4 920 - 5.81, say 6, shop rivets. More rivets 
than this will be required by practical considerations, as indicated by 



Article 54 of the Specifications or in order to make the spacing in the 
stiffener angle the same as that in the other stiffeners. This detail 
is to be left to the draftsman, the required number only being put 
on the stress sheet 

In case | of the gross area of the web is considered as efficient 
flange area, then provision must be made in the splice for the bending 
moment which the web takes. A very economical and efficient splice 
is shown in Fig. 1GO. The horizontal plates take the stress due to the 
moment, and the vertical plates take the stress due to the shear. 

The web equivalent is 3.47 square inches and the total 
moment is 1 615 000 pound-feet, which is composed of 275 000 pound- 

o o 


o o ' o 

O O O I 


.0 \ 

o~o~ o" 

oooo / 








^ 6 Shop/ 





I, M Shop) 


6 o oo 


^oV 1 ] 



0000 I 

1 o o o 

o o o / 

o o 


o o / 


Fig. 160. Splice Consisting of Vertical and Horizontal Plates. 

feet due to dead load and 1 340 000 pound-feet due to live load. 
Therefore that proportion of the 3.47 which is taken up by the dead 
load is: 

27.5 000 

X 3.47 = 0.59 square inch; 

1 615 000 

and that proportion taken up by the live load is: 
3.47 = 2.88 square inches. 

The equivalent flange area is assumed to act at the center of 
gravity of the flange ; and the bending moments equivalent to the 
above areas are, for dead load; 

0.55) X 20000 X 72.554 = 850 000 pound-inches; 
and for live load : 

2.88 X 10000 X 72.554 = 2 090 000 pound-inches. 



These bending moments must be taken up by the horizontal splice 
plates The stresses in these plates (see Fig. 160) are, for dead load: 

"kjTojT = l^ "80 pounds; 

and for live load, 

2 090 000 _ 
54 . 25 

While the allowable unit-stresses are a maximum at the center 
of gravity of the flange and are those given by the Specifications, they 
decrease rapidly towards the center of the girder, being zero at the 
neutral axis of the entire section. The unit allowable stress at the 
center of the horizontal plates will not be so great as the maximum 
allowable, but will be proportional to the distance from center (see 
Fig. 160). The horizontal plates 'will be taken 8 inches in width. 
The unit-stresses are easily determined by means of the similar 
triangles oab and oab'. The dead-load stress is determined from the 

proportion : 


20 OOP = 72.25' 

and is 14 950 pounds. For live load, the unit-stress will be one-half 
of this amount, or 7 475 pounds. 

The area required for this plate is, for dead load, -^^~ ; = 1 .05 

OO Q() 

square inches, and for live load * ==5.16 square inches, making 

/ 4 /O 

a total of 6.21 square inches for both plates. Assuming two rivet- 
holes out of the section, the net width is 8 2 ( + ) = 6 inches; 

and the required thickness for one plate is ~ = 0.52, say yV 

The joint will be weakest in bearing in the |-inch web. The 
number of rivets required is : 

15670+ 38500 


= 11 shop rivets. 

The design of the shear plate is as follows : The shear is 28 CGO 

oo f*(\(\ 

pounds, and the required area is =3.18 square inches. As 


the length of the plate is 46 j inches, the required thickness is , - 

2 X 46. .20 



= 0.034 inch, but on account of the Specifications it cannot be less 
than f inch thick. It will, however, be made i 9 ,. inch thick, since 
it will then fill out even with the horizontal tension plates and no 
filler will be required. Bearing in the web plate decides the number 
of rivets, which is: 


^4920 = sh P nVetS - 

The width of this shear plate should be, as before, 10 inches. The 
same conditions limiting the spacing of the rivets apply here as in the 
case where the splice was designed for shear only. The length of the 
horizontal plates should be sufficient to get in all the rivets, and this 
is a detail which is left to the judgment of the draftsman. 



1. A plate- 
girder is 87 ft. 9 in. 
center to center of 
end bearings. The 
dead-load moment 

is 9125000 pound- .* f 

inches, and the live- i .V ' 

load moment is } 

38 205 000 pound- 
inches, the total 
shear at the sec- 
tion being 202 700 
pounds. The web is 
90 by /..--inch, arid 
the flange angles are F1 * 1GK Proportions oMjtay^Aggto* Hearing Plate, and 

6 by 6 by -inch. 

Design the web splice when no part of the web is considered as taking 

bending moment. 

2. For the girder of Problem 1, above, design the splice when i of the 
gross area of the web is considered as effective flange area. 

77. The Bearings. Articles 113 to 119 of the Specifications 
should be carefully studied before proceeding; also Article 87. Article 
70 should be referred to, and the remarks there made about bearings 
should be read. In case the length of bearing is such as to allow a 
simple |-inch plate, care must be taken that the bearing plate does 
not extend past the flange angles more than 2 inches, or that the 
masonry plate does not extend past the tearing plate over 2 inches. 
Reference to "Steel Construction," Part II, p. 96, to Fig. 161, 




and to the discussion which follows, will explain the reason of this. 
M = = 250 x I X -i; 

12 12 : 


1X9 . 
16X 12' 

and as s = 10 000, 

250 JL _ 10 OOOJ<_9 
2 16 X 12 ' 

from which, 

I = 1.94, say 2 inches. 

In case it is desirable to have a simple masonry plate instead of a 
cast-steel pedestal, and to have the plate extend over the sides of the 

angles a distance 
greater than 2 in- 
ches, then some 
arrangement must 
be made for sup- 
porting the pro- 
jecting portion. 
Fig. 162 shows one 
of the methods 
most commonly 
used. Notwith- 
standing the brac- 
ing of the gusset 
plates, the mason- 
ry plate is not ade- 
quately supported, 
the greater pro- 
portion of the 
stress coming 
upon the ends. 

The disadvan- 
tage of having the 
masonry plate too 
long is plainly 

shown by Fig. 163. Here the girder is shown deflected under a live 
load, the rear end of the plate being tilted up and the greater part 

Fig. 162. Arrangement where Masonry Plate is Used instead 
of Cast-Steel Pedestal. 




of the pressure coming upon the forward end. The use of this style 
of plate is not to be recommended 
for spans over 40 feet. 

The design for the bearing 
of the girder under consideration 
will now be made. The total 
reaction of one girder must now 
be computed. This will be due 
to the weight of the steel in the 
girder, to the weight of the 
track, and to the reaction pro- 
duced by the E 40 loading when 
wheel 2 is directly over the end 
support. This reaction is : 

Fig. 163. Effect of Having Masonry Plate 
Too I,ong. 

Weight of Steel, 

(123.5 + 10 X 61.75) 61.75 


Weight of Track, - - (61 . 75 + 1 . 75) * 
Reaction Due to Engine Loading 


1 1 430 pounds 

6350 " 
= 99700 " 
= 117 480 pounds. 

. 117480 
ihe square inches ot bearing surlace required is ^ 


and, as the length is 1 foot 9 inches, or 21 inches, the total width of the 

cast-steel pedestal will be = 22.4, say 23 inches, or 1 foot 11 


A bearing plate must be riveted to the lower flange where it 
rests upon the pedestal. The pedestal must be so constructed as to 
allow this bearing plate to set in it. Hand-holes should be provided 
in the casting in order to allow the bolts which connect the casting to 
the girder to be inserted. These bolts should be at least f inch in 
diameter. Anchor bolts inch thick and at least 8 inches long should 
be provided and fox-bolted to the masonry. The thickness of the 
metal in all parts of the casting should be at least li inches. The 
details of the pedestal are given in Fig. 164, the length of the bearing 
being made 12 inches so as to allow one rivet to be driven in the 
flange angle in the space between the end stiffeners. 

Allowance should be made for a variation of 150 degrees in tem- 




perature. The coefficient of expansion for steel per unit of length is 
O.OOOOOG5, and the amount of expansion for 150 degrees of tempera- 
ture will be: 

0.0000065 X (61 ft. 9 in.) X 150 = 0.06 foot. 

This is about f inch, and therefore the holes in the flanges at one end 
of the girder should be made oblong and long enough to allow the 

Fig. 1C4. Side and End Elevations Showing Construction of Pedestal and Connection 
to Bearing Plate. 

girder to move f inch, or f inch either backward or forward from a 
central position. In determining the length of this slotted hole (see 
Fig. 105), it must l)e noted that the --inch bolt takes up part of this 
hole, and therefore its length should l>e I + f = say If inches. The 
width of the hole should l>e sufficient to allow for any over-run in the 
diameter of the bolt. It should be at least 1 ] inches wide. 



1. Determine the distance center 
to center of bearings, and the size 
of the masonry plate, for a plate- 
girder of 40-foot span under coping, 
the loading being E 40. 

ANS. 41 ft. 4 in.; 350 square inches. 
(NOTE Interpolate values in Ta- 
ble I, Cooper, p. 30.) 

2. If the girder span is 78 feet 
under coping, and the loading E 40, 

determine the maximum end reaction and the width of the masonry 
plate. Axs. 147 130 pounds; 24 J inches. 

78. The Stress Sheet. Plate II (p. 1 72) shows the stress sheet for 

Fig. 165. Slotted Bolt-Hole in Flange at 
End of Girder to Allow for Contraction and 
Expansion Due to Temperature Changes. 



the girder which has just been designed. It represents the best modern 
practice in that it gives, in addition to the sizes of all the sections, 
the curves for the maximum shears and moments, the rivet-spacing 
curve, and the number of rivets required in the different parts of the 
structure. This general form has been adopted by one of the largest 
bridge corporations in this country, and is to be recommended since 
it gives the" draftsman all necessary data and thus prevents the loss 
of time by an inexperienced man in recomputing certain results. The 
results just referred to are the shears, the moments, the rivet spacing, 
and the number of rivets required in the various parts. Formerly it 
was not customary to give this information on the stress sheet, and 
the draftsman was therefore required to do all this computation which 
had previously been worked out by the designer but had not been 
placed on the stress sheet in available form, and thus unnecessary loss 
of time resulted. 


79. The Masonry Plan. The same remarks which are made 
in Article 67 apply here. In this case the length of the masonry plate 
is usually determined by considerations relative to the number and 
length of the rollers in the bearing, and not by the bearing per square 
inch upon the masonry, the size of the plate as determined by the 
above considerations being usually much larger than if it had been 
determined by the unit bearing stress. A preliminary design of the 
masonry plate is usually made in a manner similar to that done in the 
case of the plate-girder; or the length of the masonry plate may be 
approximately determined from the following: 




100 feet 

23 inches 

23 inches 

125 " 

26 " 


150 " 

28 " 


175 " 

31 " 


200 " 

35 " 


The above masonry plates are for single-track bridges, with or with- 
out end floor-beams, the length being the same in either case. 



80. Determination of the Span. The determination of the span 
is made in exactly the same manner as described in Article 68. Care 
should be taken, in case end floor-beams are not used, to allow for 
the pedestal stones, which are square stones resting directly upon the 
bridge seat, and upon the top of which rest the masonry plates of the 
stringers. Their height must, of course, be such as to keep the 
stringers level. In case these stones are used, their size must be 
determined ; and if it is greater than that of the bearing or masonry 
plates, then their size determines the width of the bridge seat and 
the span center to center of bearings. 

81. The Ties. In the design of the ties, as well as in all the 
design which follows, the Specifications of the American Railway 


Fig. 166. Spacing of Stringers and Rails, and Position of Loads. 

Engineering & Maintenance of Way Association will be followed. 
Whenever reference is made to these specifications, the number of 
the article will be enclosed in parentheses, as "(5)," which signifies 
that Article 5 of the Specifications is to be referred to. 

The stringers in the bridge in question will be taken 6 ft. 6 in. 
center to center. The maximum loading (7) is such as to bring 
8 333 pounds on one tie, and to this must be added 100 per cent for 
impact, making a total of 16 667 pounds. In order to illustrate the 
method of assuming the distance, center to center of rails, as 5 feet, 
that distance will be used in this case. The maximum moment will 
then be 9 X 16 667 = say, 150 000 pound-inches. The size of the 
tie will be determined as in Article 71, the allowable unit-stress 
being 2000 pounds per square inch (5). If a 7 by 9-inch tie is 
used, the unit-stress will be 1 590 pounds. If a 6 by 8-inch tie is 
vised, the unit-stress will be 2 340 pounds. It is evident that a 7 
by 9-inch tie must be used. See Fig. 166 for spacing of stringers and 



rails, and for position of the loads. Note that, although impact 
is taken into account in this case, the size of the tie is the same as 
that designed for the plate-girder, although the unit allowable stress 
also differs. 

82. The Stringers. The width, center to center of trusses, will 
be assumed as 17 feet, since this is sufficient to clear the clearance 
diagram in cases of single-track bridges of spans less than 250 feet. 

The span which is to be designed in the following articles is a 
through-Pratt with 7 panels of 21 feet each, making a total span, center 
to center of bearings, of 147 feet inches. See Plate III (p. 251) . Rivets 
| inch in diameter will be used throughout, except in channel flanges. 

The length of the stringers end to end will be 21 feet, and accord- 
ing to Cooper's Specifications; p. 32, the maximum moment for the 
live load will be 226000 pound-feet per rail. The coefficient of 

impact (9) will be ( AA) = 0.935, and therefore the moment 

due to impact will be 0.935 X 226000 X 12 = 2535000 pound- 
inches, making a total of 5 247 000 pound-inches due to live load. 

The section modulus for any particular beam is equal to the 
bending moment divided by the unit-stress, and this is equal to the 
moment of inertia divided by one-half the depth of the beam. This 
latter quantity is constant for any given beam, and for I-beams may 
be found in column 11, Carnegie Handbook, p. 98. 

On account of the cheapness of I-beams, they will be used for 
stringers in this bridge; and sufficiently heavy shelf angles will be used 
to take up any distorting influences due to the eccentric connections 
which are unavoidable in this case. In case an I-beam had not been 
decided upon, the stringers would have been small plate-girders with 
a span of 21 feet and depth according to formula given. They would 
have been computed in exactly the same manner as a plate-girder 
span of 21 feet center to center of bearings. 

Since the dead-load moment cannot be determined until the size 
of the stringer is known, an approximate design must first be made 
by using the live-load bending moment alone; and then, with the 
size determined in this manner, the extra section modulus required 
for the dead-load moment due to the weight of the beam and the 
track must be computed. If this extra section modulus, added to the 
one previously determined, is greater than that given by the beam in 



question, a larger size beam must be used and a recomputation made. 

5 247 000 
The section modulus (17) required for live load only is 

16 000 

= 327.9. As this is too large for one beam, two beams will be used, 
thus giving a required section modulus of 164 for one beam. Two 
24-inch 80-pound I-beams will be used, giving a total section modulus 
of 2 X 174 - 348. 

Assuming the rails and ties to weigh 400 pounds per linear foot, 

the dead load per linear foot per stringer is 80 + - = 180 pounds. 

The dead-load moment is therefore 18Q X 21 X 21 X 12 = ng ^ 


pound-inches. This requires an additional section modulus of 
1 1 q f]f)f) 

= 7 .45. This, added to the 164 as determined above, makes 
lo 000 

a total required section modulus of 171.25, which, being less than 
174 (which is that for one I-beam), indicates that the above chosen 
beam is sufficient in strength, and it will therefore be used. 

The number of rivets in the end connections will now be deter- 
mined. The total end reaction for one I-beam is equal to the weight 
of one-half the beam, one-eighth the track in the panel, and one-half 
the maximum live-load reaction for one rail. These quantities are: 

Live-Load Reaction = - = 25 700 pounds. 

Impact = 25 700 X 0.935. =24 030 " 

Weight of Track = 4 ~ X - 2 -J- = 530 " 

Weight of Stringer = -- X 80 = 840 " 

Total = 51 100 pounds. 

The coefficient of impact is that for a loaded length of 21 feet. 

From p. 177, Carnegie Handbook, sixth column, it is seen that 
the longest connection angle which can be used with a 24-inch I-beam 
is 20| , say 20 inches. In this case the thickness of the connection 

angles must be - =0.23 inch; but according to (36), f 

10 000 X 2.2 

inch will be used. The angles chosen will be 6 by 3| by f-inch. 
The 6-inch leg will be placed against the web of the floor-beam in 
order to allow for sufficient room for rivets to be driven. 



The rivets will tend to shear off at places between the webs of 
the stringer and floor-beam and the connection angles. They will 
also tend to tear out of the web of the stringer and out of the web of 
the floor-beam. As the thickness of this latter is not known, the 
determination of the rivets for this condition will be made under the 
next article. The bearing value of a f-inch rivet in a $-inch plate 

(19) is I X i X 24 000 = 10 500 pounds, and therefore 51 - 11 - = 5 

10 500 

shop rivets are required in bearing in the web of the stringer. The 
value of a |-inch shop rivet in single shear (18) is 0.6013 X 12 000 = 
7 220 pounds, and the number of rivets required to prevent shearing 

between the connection angles and the webs is ~y^7r = 7 shop rivets. 
The value of a |-inch field rivet in single shear (18) is . 6013 X 10 000 

= 6 013 pounds, and therefore = 9 field rivets are required to 

6 01 o 

connect the connection angle to the web of the floor-beam. As men- 
tioned above, the number of rivets 
in bearing in the web of the floor- 

beam will be determined in the 
next article; and if the number 
required for bearing is greater 
than 9, then that number must 
be used instead of 9. Fig. 167 
shows the connection of the 
stringer to the floor-beam web, 
and also the number of rivets as 
determined above, in their proper 

9 Field 

! I-bea 

Floor 6 earn Web 

/I-beam Web 


positions. The distance between Fig. 167. Connection of Stringer to Floor- 
, , . Beam Web; also Number of Rivets. 

the webs ol the stringers must 

be such as to prevent their flanges from touching at the top. 

The stringers should be connected at the bottom by a system of 
lateral bracing of the Warren type. The size of these angles cannot 
be determined by theoretical considerations, but is usually chosen to 
be 3 i by 3 J by f-in. See Plate II (p. 172) for the general arrangement 
of this bracing. 

83. The Floor=Beams. All floor-beams should be of sufficient 
depth to allow the use of small-legged connection angles at the ends 


where they join the end-posts. The thickness of the web should also 
be greater than that which is theoretically computed, in. order that 
sufficient bearing may be given so that the rivets for the stringer 
connections will not require the stringers to be of too great a depth. 
The depth of the floor-beam will, of course, vary somewhat with the 
length of the panel and with the loading, but should not be less than 
36 inches in any case. A considerable variation in the depth will 
not affect the weight of the floor-beam or the bridge to an appreciable 
extent. A good plan is not to exceed a depth of 5 feet, with panel 
lengths of 25 feet and E 50 loading. In this bridge the depth of all 
intermediate floor-beams will be taken as 48 inches. It is good 
practice not to consider -| the web area when designing flanges of 
floor-beams and stringers, and the design here given does not consider 
the web as taking any bending moment. 

The design of an intermediate floor-beam will now be made. 
The loads for which it is designed are the floor-beam reaction due to 
the live load (see Cooper,p. 32), the floor-beam reaction due to impact, 
the dead weight of the stringers and track, and the weight of the beam 
itself. The latter .weight is distributed uniformly over the entire 
length of the beam, and the other loads act as concentrated loads 
spaced 6 feet 6 inches apart at equal distances from the center. 
The computation of the concentrated loads is as follows: 

Live Load= 68 000 pounds 


Impact = 68 000 X ( - ) 60 500 " 

o / ~r oUU 

Dead Load of Stringer = 2 ( 21 x ^ X 8 ) . . 3 320 " 

Dead Load of Track = -- X 21 4200- " 

Total = 136 020 pounds. 

The moment at points under the loads (see Fig. 168) is 136 020 
X 5.25 X 12 - 8 575 000 pound-inches. This is due to the con- 
centrated loads only. The weight of one floor-beam may be approxi- 
mately determined by the same formula as used to determine the 
weight of plate-girder spans; only, in place of the length of the span, 
the length of the panel must be substituted. The total weight of the 
above floor-beam, then, is: 

TF= 0.45 X (123.5 +10 X 21) X 21 = 3 160 pounds. 




The dead-load moment at the center duo to this weight will be: 
Wl _ 3 160 x 17 x 12 
8 8 

80 700 pound-inches, 

making a total moment at the center of the beam of 8 575 000 + 80 700 






5'- 3" ( "-e" I 5-3" 

, 7 '- *l 

Fig. 168. Diagram Showing Loads on Floor-Beam. 

= 8 655 700 pound-inches. Note that the dead-load moment at the 

center of the beam is added to the concentrated-load moment at the 

point where the concentrated load is applied. This will give the 

total moment at the center of the beam as shown by Fig. 169, since 

the concentrated-load moment, is 

constant between the points of 

application. The end shear is 

readily computed to be 136020 

+ 1 580 - 137 600 pounds. The 

curves of moments and shears are 

shown in Fig. 169. 

The total depth of the floor- 
beam, back to back of angles, is 

48i inches ; and the effective depth Fig m ghear and Moment Diagram 
will, for approximate computa- 
tion of the flange area, be taken as somewhat less, say 44^ inches, 
since the flange angles will probably be 6 by 6-inch and the center 
of gravity of most of these angles lies about If inches from 

the back. The approximate flange stress is 

pounds, and the required net area (17) 



= 194 500 
194 500 




= 12.2 square inches. In assuming the size of the angle, it is to be 
remembered that when, as in this case, no cover-plates are used, no 
rivet-holes will be taken out of the top flange, and only one rivet- 
hole will be taken out of the vertical flange. 

Two 6 by 6 by f-inch angles give a gross area of 7.11 square 
inches each, and a net section of 7 . 1 1 . 625 = 6 . 485 square inches 
each, or 12 . 97 square inches net for both. As this is near the required 
area, these angles will be taken; and a recomputation will now be 
made with the actual effective depth, in order to see if sufficient 
variation in the areas occurs to require another angle to be taken. 
The actual effective depth is now 48^ 2 X 1.84 = 44. 57 inches; 
and making computations with this, it will be found that a net area 
of 12.10 square inches is required. As this is practically the same 

as was determined at first, no 
change will be made in the size 
of the angle. 

The web is to be designed 
for a total shear of 137600 
pounds. The required area (18) is 

1QOOO = 13.76 square inches, 

and the required thickness is 
l -^ = 0.286 inch; but on ac- 

Fig.170. Calculation of Number of Rivets COUllt f tlle Specifications (36), 

through aXFTo t orBea n m le we b fStringer t inch must be used. The web 

will accordingly be 48 by f-inch. 

The determination of the number of rivets which go through the 
connection angle of the stringer and the web of the floor-beam can 
now be made. The value of a f-inch field rivet in bearing in a f-inch 
plate (19) is f X f X 20 000 = 6 560 pounds, and the total number 

required in one connection angle will be - =11 field rivets 

2t /\ o Oou 

(see Fig. 170). 

The pitch of the rivets in the flange can in this case be 
determined by the use of the formula: 





Since the flange is of the same cross-section throughout, the value 
of the effective depth will not change, and it can therefore be used in 
the above equation instead of considering the value of the distance 
between rivet lines. The shear being practically constant from the 
connection of the stringers to the end of the floor-beams, the rivet 
spacing will be constant in this distance. It will be: 

7880 X 44 . 57 

137600 -2.51 inches, 

the value of a f-inch shop rivet in bearing in the f-inch web being 
| X f X 24 000 = 7 880 pounds. This is seen to be less than 2f 
inches ; but, as the angles have 6-inch legs, this spacing can be used in 
a horizontal direction ; and the distance from center to center of rivets, 
which will be placed in rows on two gauge lines, will still be greater 
than 2f inches. 

The shear between the stringer connections is practically zero, 
and therefore the spacing will be very large. Being over 6 inches, it 
will be subject to (37). 

The connection angles at the ends of the floor-beams will be 
taken as 6 by 3^ by f-inch, the 6-inch legs being against the web of 
the floor-beam. The other legs are chosen small in order that they 
may fit into the channels which will very likely be required for the 
posts; and according to the sixth column, p. 183, Carnegie Handbook, 
only 8} inches is available for this purpose. This 8| inches is taken 
from a 10-inch channel, since this is the smallest channel that can be 
used which will give room for connection and yet be in accordance 
with the Specifications. This is due to the fact that its web (36) is 
greater than f inch. The rivets which connect the end angles to the 
floor-beam web are shop rivets, and those which connect the end 
angles to the posts are field rivets. Since the size of the post is not 
known, the thickness of its metal, of course, cannot be used, and 
therefore the number of rivets required in bearing in the post cannot 
be determined at this time. 

The number of shop rivets required through the end angles 
and the floor-beam web is governed by the bearing of the rivets in the 
f-inch web of the floor-beam. The value of a f-inch shop rivet in 
bearing in a f-inch web (19), as has just been computed, is equal to 

7 880 pounds, and the number of rivets required is =?= 18. 




bers may be 

The number of field rivets required in single shear to connect 
the end angles with the posts is ^ = 23. An even number of 

rivets will, of course, have to be used, one-half going into each 
of the 3i-inch legs. See Fig 171 for the position and the number 
of rivets. It must be remembered that more than these num- 
used by the draftsman on account of rivet spa- 
cing which may be required by 
conditions other than those of 

The design of the end floor- 
beam is somewhat different from 
that of the intermediate floor- 
beams in that the load which 
comes upon it is considerably 
lighter, since this floor-beam 
takes the dead load of only one- 
half the panel and the live load 
due to the maximum end reaction 

Channel of 

Web ot Floor 

Fig. 171. Position and Number of Rivets 
to Connect End Angles with Posts. 

for a stringer length instead of 
the floor-beam reaction for the 
stringer length (see Cooper, p. 
The maximum end shear is computed as follows: 

End Shear for 21-foot Span 51 400 pounds. 

300 \ 

Impact = 51 400 X 


Dead Load of Stringers = 

Dead Load of Track = - 

- 300 / ' ' 

80 X 2 X 21 


400 X 21 

. 48 000 


2 100 

Total 103 180 pounds. 

The maximum moment due to the above load is 103 ISO X 5.25 
X 12 = 6 500 000 pound-inches. The w r eight of the beam may be 
assumed as 3 160 pounds. This is the same as was computed for the 
intermediate floor-beam, but will be used for this beam, since the size 
of the web will be the same as in the others; and, although the flange 
area will be less, the end connections will be somewhat heavier owing 
to the connection of the beam to the end-post and the roller bearing, 



and this additional weight will cause the total weight of the end 
floor-beam to be about the same as that of the intermediate ones. 
The total moment at the center will then be 6 500 000 + 80 700 - 
6 580 700 pound-inches. 

The depth of the end floor-beam will be somewhat greater than 
the depth of the intermediate floor-beams. This is due to the fact 
that it extends downward a greater distance, resting upon the bearing 
plate, which comes directly upon the top of the rollers. The exact 
depth cannot, of course, be determined until after the roller bearings 
are designed; but it may be safely assumed as four or five inches 
deeper than the intermediate floor-beams, and in case this is riot 
enough, the draftsman can easily fill in the remaining distance with 
filler plates, as this distance will not be very great. In case this depth 
is too great, the flange angles may be bent upward at the end, or a 
re-design may be made. 

The depth will be assumed as 52 inches in this case. The 
effective depth will be assumed as 48 inches, and this gives an approxi- 
mate flange stress of 

6580TOO =137000poundS) 

and an approximate net flange area required of 

137 000 

1fi -_- = 8.57 square inches. 

A 6 by 6 by -^g-inch angle gives a gross area of 5.06 square 
inches, and a net area of -5. 06 (f + |) r \ = 4.62 square inches. 
A recomputation with the true effective depth requires 8.42 square 
inches net. Two of these angles give 9.24 square inches; and as this 
coincides very closely with the required area, it will be used; The 
size of the web plate is 52 by f-inch. 

The pitch or spacing of rivets in the flanges is : 

7 880 X 48 

~ 104 7CO = ' mches - 

The maximum end shear as above computed is taken by two 
stringers; and therefore the number of rivets required in bearing to 
form the connection between the stringers, connection angles, and 
the floor-beam web is, for each angle: 

104 760 

.TT = 8 field rivets. 

2 X 6 560 



The value 6 560 in the above equation is the value of a |-inch field 
rivet in bearing in the f-inch web. 

The number of rivets required in the end angles on the floor- 
beam is : 

104 760 

- y ^ r = 14 shop rivets. 

These rivets go through the web of the floor-beam. The connection 
of the floor-beam to the end-post is made by means of field rivets and 
a large gusset plate. This gusset plate is usually | inch in thickness. 
The number of rivets through the end connection angles and this 
gusset plate is governed by single shear, since the rivets will shear off 
between the angles and the gusset plate before they will tear out of 
the gusset plate, as the value of a rivet is greater in bearing than in 
shear. The number required is: 
104 760 


= 18 field rivets. 

The general arrangement of the intermediate floor-beams is 
shown in Fig. 172. The ends of the lower flange are bent up as 
shown, in order to allow the I-bar heads or any other section of the 
lower chord to have clearance. This makes it necessary for the floor- 
beam web to be spliced at the ends, as shown. The distance which 
this plate should extend above the floor-beam proper depends upon 
the distance which the lower chord is bent up. In any case the length 
of the connection on the post should be at least equal to the depth 
of the floor-beam. Two splice plates, one on either side of the web, 
are placed here in a manner similar to that of a splice as designed in 
the plate-girder when shear only was considered. Here shear only is 
considered, and the number of rivets which must be on each side of 
the splice will be: 

137 640 
7880 = 18 shop rivets. 

The 7 880 which occurs in the above equation is the value of a |-inch 
rivet in bearing in a f-inch plate (19). Inspection of Plate II (p. 
172) will make this design clearer. Plate II also shows the shape of 
the end floor-beams. 

The small shelf angle shown in Fig. 172 should have sufficient 
rivets to prevent any twist of the stringers due to their being con- 
nected on one side of their web only. This number is a matter of 




judgment. Experience seems to indicate that enough rivets to take 
up one-third of the total reaction of the stringers will be sufficient. 
This will require shop rivets, and the number will be : 

103 180 
- = 5 shop rivets in single shear. 

84. The Tension Members. Tension members usually consist 
of long, thin, flat plates with circular heads forged upon their ends. 

Fig. 17Z. General Arrangement of Rivets, Splices, Connections, etc., for Intermediate 

These circular heads have holes punched through their centers and 
then very carefully bored. Through these holes are run cylindrical 
bars of steel called pins. These pins connect them with other mem- 
bers of the truss. See Carnegie Handbook, p. 212, for table of I-bars. 
The I-bars given are standard I-bars; and while departures from these 
widths and minimum thicknesses may be made, it may be done only 
at great cost to the purchaser. Note that there are no standard 
9-inch I-bars. The thicknesses given are the minimum thicknesses 



for that width of bar, and do not indicate that thicker bars of that 
width cannot be obtained; but on the contrary thicker bars of that 
width can be obtained, and this should be done, the minimum thick- 
ness as given in the table being avoided if possible. 

It has been found that bars which have a ratio of thickness to 
width of about one-sixth give good service and are easy to forge. 
This relation gives us a rough guide which will enable us to determine 
the approximate width and thickness of any bar of a given area. 
Once the approximate dimensions are determined, the actual dimen- 
sions can be chosen from the market sizes of the material (see Car- 
negie Handbook, pp. 245 to 250). 

An expression for the approximate de,pth of the bar will now be 
derived by using the above relation. 

Let A = Area of bar, in square inches; 
d = Width of bar, in inches; 
t = Thickness. of bar, in inches. 

td = A; 

Substituting the value of t in the expression for A, there results: 

d = VGA. 

The stresses in all the members in the truss under consideration 
are computed by the method described in Part I, and are placed on 
the stress sheet, Plate III (p. 251). In the succeeding design, the 
student should obtain his stresses from Plate III without his attention 
being again called to the matter. 

Table XXIV gives the tension members and their dead-load, 
live-load, impact, total, and unit stresses (15), together with the 
required area, the number of bars, the approximate depth of bars, 
and the final sizes used. 

The first seven columns in Table XXIV are self-explanatory. 
The number of bars to be used in any particular case is a matter of 
judgment. One fast rule is that an even number of bars should 
always be used, except in the case of counters, where one is permis- 
sible. This is due to the fact that the placing of one of the main 
members in the center of the pin would create a large moment, and 




i i 





* NOTE All areas in square inches, and all dimensions in inches. 
Observe that the above table is not completely filled out with respect to the first two members given. This is on ac- 
count of the requirements of the Specifications (80). 

therefore, an ex- 
cessively large pin 
would be required, 
and accordingly a 
very large head on 
the I-bar in pro- 
portion to its width 
all of which are 
very undesirable 
and costly. In gen- 
eral the number of 
I-bars should be as 
small as possible, 
and they should be 
so chosen that the 
widths of the chord 
members increase 
from the ends to- 
ward the center of 
the truss, and the 
widths of the diago- 
nals decrease from 
the ends toward the 
center of the truss. 
The area of one 
bar is obtained by 
dividing the total 
stress by the num- 
ber of bars and also 
by the allowable 
unit-stress. Thus, 
for the member 
LJJ V for example, 
the required area 
of one bar is : 
295 100 



. <*oo*o *. 

'. '. X X X X X X 




: ;S ;-8SS 


: :8SSSS 

; 1 p> i-I >o *o bi oo 


; ; (N ^H (M Tt< Tfl 01 











ESSES (in poun 



;3 2 >-i 53 


rH CO t~- .' Tf< 00 <N .' 

<* rt T}< . C^ CO 00 . 






. 9 22 
2 X 16 000 

square inches. 



The approximate depth of this bar is determined by taking the square 
root of 6 times the area as above determined. It is: 

d = V 6 X 9 . 22 = 7 . 44 inches. 

As this is nearer 7 than 8 inches, a 7-inch bar will be chosen; and 
looking in the first column, Carnegie Handbook, p. 248, for an area 
which will be equal to or in excess of 9.22, it is found that a 
If -inch bar satisfies this condition, and therefore the section of this 
member consists of two bars 7 by If inches. 

According to (80), the first two sections for the lower chord are 
to be made of built-up members. This requires that instead of 
I-bars they are to be made of angles and plates, or, in case the stress is 
light, of channels. The depth of the section is limited by the size 
of the greatest I-bar head. As the diameter of the I-bar head depends 
upon the size of the pin, it cannot of course be determined accurately 
before the pin is designed. It is customary to assume the largest 
head, and to design the section so as to clear this. The size of the 
largest head for bars of given width is given in the Carnegie Hand- 
book, p. 212. 

The design of the member L L 2 will depend upon the size of 
the largest head of the 7-inch I-bar of the member L\L r This is 17 \ 
inches; and in order that the head may have some clearance, it will 
be necessary to add \ inch to the top and the bottom, making a total 
of 18A inches. Since the flange angle, as in the case of plate-girders, 
will extend over the plate about \ inch, the plate itself may be 18 
inches wide and still give sufficient clearance. 

The total stress is 234200 pounds, and the allowable unit- 
stress (15) is 16 000 pounds per square inch. The required net 
area, then, is: 

234 200 
160QO = 14. 64 square inches. 

According to the Specifications, the thickness of the plate cannot be 
less than f inch. The gross area of two 18 by f-inch plates is 13.5 
square inches, and the gross area of four 3^ by 3^ by f-inch angles, 
which are assumed to be sufficient, is 9 . 92 square inches, thus making 
a total gross area of 23 . 42 square inches. If 5 rivet-holes are assumed 
to be taken out of each web, and one rivet-hole taken out of each 
angle, this will require a certain number of square inches to be 





deducted from the section, and this is computed as follows: 

Out of webs, 2 X 5 (| + ) X -jj- = 3.75 sq. in. 
Out of angles, 4 (J + i) X f . . = 1 .50 " " 

Total = 5 . 25 sq. in. 

The net area of the section is now determined to be 23 .42 5 .25 = 
18.17 square inches. This is somewhat greater than the required 
net area, but must be used, for according to (39), these are the smallest 
and thinnest angles that may be used. 

Figs. 173 and 174 show the cross-section and the general detail 
at L 2 . The width of the member cannot be determined until after 
the section of the end-post is 
computed, since it must fit inside 
of the end-post, the horizontal 
legs of the angles being cut off to 
allow this. The end-post, Arti- 
cle 87, is 14 1 inches inside. If it 
is assumed that all the pin-plates 
on the end-post are placed on 
the outside, and all those at L 
on L L 2 are on the inside, then 
the width of L L 2 , back to back 
of plates, must be 14 - (2 X 2 
+ 2 X 1) = 121- inches or less, 
J-inch clearance being allowed 
between the sides of the angles and the web plates of the end-post 
(see Fig. 173). 

The total net section through the pin-hole at L 2 (26) must be 
1-|- X 18.17 = 22. 7 square inches, or 11.35 square inches for one 
side. The plate which is to increase the section must be on the out- 
side, since the intermediate post U 2 L 2 and the two I-bars of member 
UiL 2 must go inside. The gross width of this plate is 1 1 } inches (see 
Fig. 174), and the net width is 2w = \\\ - 5 = 6J inches. The 
net area through the pin is: 

Two 3^ by 3 by f-in. angles = 4 . 96 square inches. 

One 18 by \-\n. plate = 9 - 5 X \ = 6 . 50 " 

Total = 11 .40 square inches 

Since this is greater than the 11 .35 required, no plate will be necessary 
to fulfil (26) in this respect. 

Fig. 173. Cross-Section Showing Construc- 
tion of Lower Chord Member. 




u\~~ iK\\ 

13 I j/*\ 



Sufficient bearing area must be provided at this point. The 
total stress is 234 200 pounds, the total bearing area required is 

234 200 
' = 9.76 square inches, and the total thickness for one side is 

^4 UUU 

: - = . 976 inches. Since the thickness of the web is $ inch, the 

pin-plates must be 0.976 - 0.50 - 0.476 inch (say \ inch) thick. 
A |-inch pin-plate must be used, and as the total thickness of the 

bearing area is now 1 .00 inch, this pin-plate will take '- 

= 58 550 pounds. The joint is weakest in shear, and will therefore 

require ---~ = 8 + (say 9) shop rivets. 

In case it is necessary to put the member U^ on the outer side 
of L L 2 , then the outer legs of the upper angles must be cut off to allow 
t/^Lj to pass. This will decrease the section by an amount (3^ f ) 
X f = 1 . 20 square inches. Considering the pin-plate, which is 
(18$- 2X3$) i=llj inches, the | inch being allowed for 
clearance between the edges of its flange angles, the total net 
section through the pin-hole on one side will be: 

One Angle 3 by 3 by f-in. = 2.48 square inches 
One Cut Angle 3 by 3 by f-in. = 1.31 " " 

One Web (18 - 5) J sq. in. =6.50 " 

One Pin-Plate (11J - 5) f sq. in. = 2.34 " 

Total = 12 . 63 square inches. 

This is greater than 11 .35 as required, and is therefore safe. 

The distance from the center of the pin to the end must now 
be determined (26). The total net section of the body of the member 
is 18.17 square inches, or 9.09 square inches for one side, and the 
thickness of the web and the pin-plate is 1 inch. The distance from 

9 09 
the pin to the end of the member is then - = 9^ inches, and the 

distance to the center of the pin is 9|- + ~ -= llf, say 12 inches (see 

Fig. 174). Rivets should be countersunk where necessary to prevent 
interference with I-bars. For signs, see "Steel Construction," Part 
III, p. 192, and Carnegie Handbook, p. 191. 




At point L of this member, the pin is 6J inches in diameter, and, 
as previously mentioned, the legs of the angles are cut (see Fig. 175). 

The total bearing area required for one side is '- = 4 . 88, and the 
required thickness is ^^- = 0. 781 inch. Subtracting the thickness 

O . -3 

of the Wnch web from this gives 0.281 inch. A pin-plate f inch 
thick must be used. 

The net area through the pin (26) must be 1 1 . 35 square inches. 


*. / 



v, //{' 



/ ^ 

e Pm Plate 

i j ^ 



i i 


Fig. 175. Elevation and Section Showing Pin Connection at End of Truss. 

This net area, remembering that the angle legs are cut and therefore 
their area is that of a bar 3^ by f-inch, computed for one side, is as 
follows : 

Two Angles, legs cut, 3 by 3J by f-in. = 2.62 square inches 

One Web 18 X i - 6 X \ sq. in. = 5.88 " 

One Pin- Plate Hi X | 6J X f sq. in. = 1.87 " " 

Total 10 . 37 square inches. 

This shows the section to be deficient, and the thickness of the pin- 
plate must be made \ inch. This gives a net area through the pin 
of 11 .62 square inches. 

The distance between rivet lines (see Fig. 173) is \1\ inches, 
and (44) the tie-plates must be 17 \ (say 18) inches long, and their 

17 25 

thickness '- = 0.346 inch (say f inch). 



The lattice bars (45) must be 2^ inches wide, and (47) must be 
double. From (45) and Table XXV, page 219, the thickness must 
be r 7 6 inch, the distance c being 17.25 X secant 45 - 2 ft. T \ in. 

The design of the hip vertical U 1 L 1 is also made in accordance 
with (80) of the Specifications. It will be assumed that the section 
consists of one 8 by f-inch plate, and four 3^ by 3| by f-inch angles, 
since this is the lightest section that may be used according to the 
Specifications, the 8-inch plate being chosen as it gives some clearance 
between the inner edges of the legs of the angles. 

The total stress in the member is 141 600 pounds, and the unit- 
stress is 16000 pounds per square inch, thus requiring a net area of 
8.85 square inches. The plate gives a 
net area of 2 . 20 square inches, and the 
four angles give a net area of 6.88 square 
inches, making a total of 9.08 square 
inches, one rivet-hole being taken out of 
each angle, and two out of the web, at 
any particular section. The net area is 
somewhat greater than that required, but 
must be used, as this is the minimum sec- 
tion allowed by the Specifications. Fig. 
176 shows a cross-section of this member 
as above determined. 

This member will be connected to the upper chord and end-post 
by means of a pin which is 6| inches in diameter, the diameter of the 
pin being determined later. The total stress is 141 600 pounds, and 
this will be taken by two plates, one on either side of the member. 
The net section of the member is 9.08 square inches, and the section 
through the pin (26) must be 25 per cent in excess of this, making a 
total of 11.35 square inches, or 5.68 square inches for each plate. 
The total width of these plates will be taken as 12 inches, and this 
(see Fig. 177) will make the required thickness: 

Fig. 176. Cross-Section of Hip 


-=0.99 inch. 

12.00 - 6.25 

Fig. 177 shows the details of these pin-plates. Since the above 
thickness is too great to be punched in one single piece, the above 
thickness will be made up of two plates, each ^ inch thick. The 
area at section A-A must be equal to that of the body of the bar. It 



is 12.00 X $ = 6.00 square inches for one side, or 12.00 square 
inches for both sides. As this is greater than the 8 .85 square inches 
as above computed, the area at A- A is sufficient, as is also the width 
of the plate, which was assumed as 12 inches. 

One of the plates will be riveted directly to the member, and the 
other will be riveted to it as a pin-plate. The section back of the pin 
(26) must be equal to the net section in the body of the member. The 
net section is 4.54 square inches for one side, and the total thickness 

Fig. 177. Connection of Hip Vertical to Upper Chord and End-Post. 

of the pin-plates is 1 . 125 inches, making the distance from the end 

4 54 

of the member to the pin '-- = 4% inches, and the distance to 
1 .125 

the center of the pin 4| + -^ = 1\ inches. 

The joint between the plates and the main member will be weak 
in shear, the rivets tending to shear off between the f -inch angles and 
the plate, and also between the two plates themselves. As each side 
takes one-half of the above stress, the number of rivets required to 
connect the plates to the main member will be: 

141 600 * 2 

= 10 shop rivets, 

and the number of rivets required to connect the inner -J-inch plate 
to the outer one which is connected to the member itself will be: 
141 600 H- 4 


5 shop rivets. 




The distance from the center of the pin to the top of the main 
part must be greater than one-half the diameter of the largest I-bar 
head that is, 17^ -j- 2 = say, 9 inches. 

At the lower end, this member is connected to the bottom chord 
by means of a couple of clip angles and four or five rivets. Only 
sufficient rivets are required to prevent the sagging of the bottom 
chord, since the floor-beam is connected to the hip vertical above the 
lower chord, and hence no stress comes on the joint at the lower end 
(see Fig. 178). 

The width of the plate has been assumed as 8 inches. This 
width is liable to be changed after the design of the intermediate 

Fig. 178. Connection of Hip Vertical to Lower Chord. 

posts has been made, since it will be economical to have all the inter- 
mediate floor-beams of the same length; and therefore the width of 
this plate will be changed so as to make the width of the hip vertical 
the same as the width of the intermediate posts. 

85. The Intermediate Posts. The post UyL 2 must be designed 
to stand a total stress of 163 600 pounds. Where possible, it is 
economical to make the intermediate posts out of channels, as this 
saves a large amount of riveting. As seen by the stress sheet, the 
length of these posts is 30.1 feet center to center of end pins. It is 

usually required that must not be more than 100, and this con- 
dition requires that the least radius of gyration cannot be less than 
30.1 X 12 


= 3.62. 



From Carnegie Handbook, p. 101, it is seen that a 12-inch 30- 
pound channel has a radius of gyration of 4.28, and will fulfil the 
conditions. The area of two of these channels is 17 . 64 square inches. 
The unit allowable stress (16) is: 

P = 16 000 - 70 X X - = 1 090 pounds per square inch. 

The required area is then determined to be 1OftQO - 16.2 square 

inches; and as this coincides very closely with the area given, these 

channels are efficient and will be used. 

Fig. 179 shows the cross-section of this post. The radius of 

gyration which was used above was the radius of gyration of the chan- 
nels about an axis perpendicular 
to their web. The radius of gyra- 
tion of the entire section about 
an axis perpendicular to the 


,5 Shop 9 Shop 

e Field 


I/- is Field 

web will be the same as that of 
one channel. In order to have 
the sections safe, the radius of 
gyration about the axis B-B must 
be equal to or greater than the 

about the axis B-B can be in- 
creased or decreased by spacing the channels. The exact distance 
which will make the two rectangular radii of gyration equal may 
be determined by the methods of "Strength of Materials," or it 
may be found in columns 14 and 15 of the Carnegie Handbook, p. 
102. For any particular case it is equal to the value given in column 
14, plus four times that given in column 15. For the channels under 
consideration, it is equal to 7.07 + 4 X 0.704 = 9. 89 inches. Any 
increase in this distance will only tend to increase the radius of 
gyration about the axis B-B, and will make the post safer about 
that axis. 

Fig. 179 shows a diaphragm. The web of this diaphragm cannot 
be less than f inch, and the size of the angles cannot be less than 3^ 
by 3 \ by f -inch, as this is the least allowed by the Specifications. The 
function of this diaphragm is to transfer one-half of the floor-beam 
reaction to the outer side of the post. The rivets which connect the 


angles to the diaphragm web are shop rivets, and (see design of floor- 

137 600 
beam) must be- - ^-^' 9 in number. The rivets which connect 

2X7 880 

the diaphragm angle with the outer channel of the post are also shop 
rivets, and are -^ - = 10 in number, 5 on each side. The 

y\ / ' ' 

same rivets which connect the floor-beam to the post go through 
the diaphragm angle on that side of the diaphragm next to the cen- 
ter of the bridge, and must therefore be field rivets and take the 

entire floor-beam reaction. These must be ----- = 23 in number, 

6 013 

12 on each side. The exact distance, back to back of the channels of 
the post, cannot be determined until after the top chord has been de- 
signed, since the post must slide up in the top chord and also leave 
room on each side for the diagonal members of the truss. The 
width is determined by the packing of the members at joint L 2 (see 
Fig. 174), and is found to be 9J inches. Since this is less than that 
required above, the post must be examined for bending about an axis 
parallel to the web of the channels. 

According to the methods of "Mechanics" and "Strength of 
Materials," with the help of the Carnegie Handbook, p. 102, the 
moment of inertia about this axis is found to be 286.42, and the ra- 
dius of gyration 3 .96. The unit allowable compressive stress is then 
computed to be 9 580 pounds per square inch, and the required area 

163 600 

_ ---- = 17.10 square inches, which, being less than 17.64, shows 

the section to be safe. 

This member is connected to the top chord at its upper end by 
a 5-inch pin. The total stress is 163 600 pounds, and the total bear- 

163 600 s 

ing area required is - = 6.8 square inches, or 6 .4 square inches 
^4 (JOU 

for each side (19). The total thickness of the bearing area for each 

side is - = 0.68 inch. The thickness of the web of a 12-inch 

30-pound channel is 0.513 inch, which leaves 0.68 - 0.513 = 0. 167 
inch as the thickness of the pin-plate, but it must be made f inch 
according to the Specifications. Fig. 180 shows the arrangement of 
the plates and the rivets. 




The sum total of the pin-plates and the channel web is 0.888 
inch, and therefore on one side the stress transferred to the pin by 

1 ^ 0.375 X 163 600 
means of the pin-plate, which is . 375 inch, is X O~QQC 

= 34 600 pounds. This plate will tend to shear off the rivets between 

34 600 
it and the channel web, and therefore = 5 shop rivets are 


The stress that is shown on the stress sheet is the stress in the 
post above the floor-beam. The stress in that part below the floor- 


l ' 




K 1P 

-- - 

--jif--- 3 







1 1 



M I 













1 1 1 l 







1 1 
















Fig. 180. Arrangement of Plates, Rivets. Pin, etc., at Connection of Intermediate Post 
to Top Chord. 

beam is equal to the vertical component of the diagonal in the panel 
ahead of the post in question. In this case it is the vertical com- 
ponent of the stress in U^, and is equal to 242 000 pounds, and this 

. 242 000 
requires a total bearing area of -^77^7^ = 10.1 square inches, and 

a total thickness of 

v V" = 1.01 inches on each side, the pin being 

5 inches in diameter. From this total thickness must be subtracted 
the thickness of the web of the channel, and this leaves 1.01 0.513 
= 0.497 inch as the total thickness of the pin-plates required. This 
shows that we must use one ^-inch plate. The total thickness of the 
bearing area is now 0.51 3 + 0.50 - 1.013 inches. 

Each plate takes a total stress of 

= 59 700 




pounds; and the joint being weak in shear, the number of rivets 

. , . , 
required will be 

59 700 

= 9 rivets in single shear. The detail will be 

similar to that in Fig. 180. 

The distance, back to back of the channels in this post, will 
probably not be greater than 12 inches, and this will make the dis- 
tance between rivet lines about 9 inches. According to (44), the end 
tie-plates must be at least 9 inches long and of course 12 inches wide. 


The thickness cannot be less than = 0.18 inch, but they will be 


made f inch (36). Between the tie-plates the channels will be con- 
nected by means of lattices. The Specifications (45) require that 

they should not be less than 1\ inches in width and (1 .414 X 9) - = 


0.318 (say f) inch in thickness. Table XXV gives the thickness of 
lacing bars for any distance between rivets. 

Thickness of Lacing Bars 



(<=40 : ^= 30 ) 


(i=^: = 45o) 





1 in. 

Oft. 10 in. 

i in. 

1 ft. 3 in. 

js in- 

1 ft. 0^ in. 

W in. 

1 ft. 6} in. 

f in. - 

ft. 3 in. 

1 in. 

1 ft. lOJt in. 

I 7 c in. 

ft. 5i in. 

T 7 in- 

2ft. 2in. 

i in. 

ft. 8 in. 

i in. 

2 ft. 6 in. 

T in - 

ft. 10^ in. 

T* in - 

2ft. 9|in. 


2 ft. 1 in. 

f in. 

3ft. Hirf. 

A width of 2} inches is chosen above, since according to Carnegie 
Handbook, p. 183, a f-inch rivet is the largest which can be used in 
the channel flange. 

The post U^ must be designed for a total stress of 87000 
pounds. It will be assumed that two 10-inch 20-pound channels 


with a radius of gyration 3 .66 and an area of 5 .80 square inches each 
will be sufficient. The length, as before, is 30.1 feet, and the unit- 
stress is: 

qrj i \/ 1 O 

P = 16 000 - 70 X - 9 080 pounds, 


The required area is = 9.60 square inches. Since the 

3 UoU 

total area of the two channels is 11 .76 square inches, and the required 
area is 9.6 square inches, it is seen that the} 7 do not coincide very 
closely. These channels, however, will be used, since the thickness 
of the web is the thinnest allowed by the Specifications, and the 
width of the channels is the smallest that can be used and still give 
sufficient room to make the connections with the end connection 
angles of the floor-beams. 

The lower end of this post also has a diaphragm which must 
transfer half of the stress to the outer channel of the post. The sides 

of the diaphragm are the same 
as in the posts previously de- 
signed; and the number of rivets 
required is computed in a simi- 
lar manner and found to be as 
indicated in -Fig. 181, which 
shows the cross-section of this 

At the upper end the bear- 
Fig. 181. Cross-Section of Intermediate . . 

Post. mg area required on one chan- 

87 1 00 

nel is- = 1.814 square inches, and the thickness required 

tL /\ <u4 UuU 

is ^ == 0.363 inch, a 5-inch pin being used. As the web of the 

channel is 0.382 inch thick, it will give sufficient bearing area 
without pin-plates. 

At the lower end, the vertical component of UJL 3 is 157 500 

pounds. The bearing area required on each side of the post is 
-i r*j KAH o oo 

= 3.28 square inches, and the thickness is = 0.66 

2i X 24 000 5 

inch. The thickness of the channel web being . 382 inch leaves . 660 

0.382 = 0.278 inch as the required thickness of the pin-plate; 



but f inch must be used, making a total thickness of 0.382 + 0.375 

= 0.757 inch. The plate will carry ~?j? X ~ = 39 000 

pounds, and this requires - = 6 shop rivets in single shear. 

The distance, back to back of channels, will be the same as in 


Fig. 182. End and Side Elevations Showing Detail of Construction at Lower End of 
Intermediate Post. 

C7 2 L 2 , and therefore the tie-plates and lacing bars will be the same. 
Fig. 182 gives a detail of the lower end of U 3 L 3 . 

86. The Top Chord. The top chords of small railway bridges 
may be made of two channels laced on their top and bottom sides. 
This is not very good practice, since it leaves the tops of the channels 
open and lets in the rain and snow, wjiich tends to deteriorate the 
joints. It is better to add a small cover-plate, even if this does give 




an excessive section. In case of stress such 
as is demanded, the chords may consist of 
two channels and a cover-plate. In this 
case it is necessary to place small pieces 
called flats upon the lower flanges of the 
channel, in order to lower the center of 
gravity of the section and to bring it near 
the center of the web. This section makes 
a very economical section in that it saves 
much riveting. On account of channels 
being made only up to 15 inches in depth, 
' the use of this section is quite limited owing 
to the fact that it is not deep enough to 
allow the I-bar heads sufficient clearance, 
for the I-bar heads in bridges of even ordi- 
nary span will exceed this amount. 

The most common section is that which 
consists of two side plates, four angles, and 
one cover-plate. Sometimes this section 
has flats placed upon the lower angle in 
order to lower the center of gravity, as ex- 
plained above. According to (33), the sec- 
tion should be as symmetrical as possible, 
and the center of gravity should lie as near 
the center of the web as is consistent with 

In case the stress is great enough to 
demand a heavier section than that above 
described, additional plates are added upon 
the sides of the original plates, and heavier 
and larger cover-plates and angles are used. 
Fig. 183 shows different types of chord sec- 

Ir addition to the cover-plate being 
designed to withstand the total stress, close 
attention must be paid to (42). This clause 
has been inserted on account of practical 
considerations, since it has been found out 


that if plates are made much thinner than the proportions here 
required, they will crumple up and fail long before the allowable 
unit of stress as computed from the formula has been reached. In 
some cases especially where the stress is light the proportions laid 
down in (42) and (36) will govern the design of the section, instead of 
the required net area as determined by the formula for the allowable 
unit compressive stress. 

The design of the first section of the top chord will now be made. 
Here, as in the case of the first sections of the lower chord, the 
diameter of the head of the greatest I-bar determines the width of the 
plates in the section The head of the 7-inch I-bar which constitutes 
the member U t L 2 is 17^ inches, and, allowing a clearance of | inch 
on either side of the head, the total depth inside the chord should be 
18| inches. As in the case of the lower chord, plates 18 inches wide 
may be used. 

The size of the angles to be chosen is a matter of judgment. 
Usually any size should be chosen at first, and the preliminary design 
will indicate at once what size should have been taken. For this case, 
3^ by 3i by f-inch will be assumed at first. 

For sections of this character, the radius of gyration is approxi- 
mately equal to 0.4A, in which h is the height, or rather the width, 
of the side plate. The approximate radius of gyration is r = 0.4 X 
18 = 7.2 inches, and the length is equal to one panel length, or 21 
feet. The allowable unit of stress (16) is: 

P = 16 000 - 70 X 21 ? X 2 12 = 13 550 pounds. 

449 500 
The required area is ^-^n = 33.2 square inches. The correct 

lo OOU 

proportion for sections of this character is that . 4 of the total area 
should be taken up by the web. The area of the web would then be 
0.4 X 33.2 = 13.28 square inches, and the thickness would be 

L 3 J^L = o.37 inch. According to this, a f-inch plate should be used, 
2 X 18 14 5 

but (42) requires that it shall be.-^- = 0.483 inch or thicker. 

Therefore an 18 by ^-inch plate must be used for the web. 

The correct proportion for sections of this character is that the 
width between plates should be about the width of the side plates. 




This will give the required width between plates equal to | X 18 = 

15.75 inches. The cover-plate (42) must not be thinner than 


the distance between the connecting rivet lines. The rivet lines are, 
in this case, 15.75 + 2 X 2 = 19.75 inches apart, and therefore the 

thickness of the cover-plate cannot be less than = 0.494 inch. 


__ Neutral Axisr-v 
Center Line of Pins.J 

The cover-plate will therefore be taken as ^ inch thick. The width 

of the cover-plate (see Fig. 184) 
must be about 15.75 + 2 X 34 
+ i = 23| inches (say 23 inches). 
The cover-plate will be taken 23 
by 4-inch. 

The center line of pins will 
be taken at the center line of the 
web, and the center of gravity of 
the section w r ill be assumed as ^ 

inch above this. In order that 

j the center of gravity may be near 

. 1 j_ that assumed, the moment of the 

Ai ' -gh* I ""! cover-plate about the assumed 

center of gravity axis should be 
about equal to the moment of the 
flats about the same axis. The 
moment of the cover-plate about the assumed axis is: 

OQ -y Q 

(9.0 - 0.5 + 0.25 + 0.25) = - 


I I 

Fig. 184. Section of Top Chord. 

and the moment of the flats about the same axis is: 
A (9.0 + 0.5 + 0.25 + 0.5) = 10.25 A, 

in which A is the area in square inches of both of the flats. Equating 
these two expressions, and solving for A, there results: 
23 X 9 

A = 

2 X 10.25 

= 10.1 square inches. 

Assuming the flats to be 4 inches wide, the thickness on each side 
will be 1 .25 inches. As this is too thick to punch, the flats on each 
side will be composed of two 4 by f-inch plates. 
The total area is: 


One cover-plate = 2.3 X i = 11.5 sq. in. 

Two web plates = 2 X IS X i =18.0 " " 
Two flats 4 X 1} =10.0 " " 

Total 39.5 sq. in. 

But the required area is 32.2 square inches, which is considerably 
less than the area above given, and which does not include the angles 
and hence we can use the smallest size angles, which are those pre- 
viously assumed. The area of each of these angles is 2.48 square 
inches, thus making the total area of the section 39 . 5 + 4 X 2 . 48 = 
49.42 square inches. This is considerably in excess of the area as 
required according to the formula for compression; but it is the least 
allowed by the Specifications. Note that this is the case where 
(42), instead of the formula for compressive stress, is the ruling factor 
in the determination of the section. 

The center of gravity of the approximate section must now be 
determined, the moment of inertia and the radius of gyration about 
the neutral axis must be computed, and the required area must be 
determined by using this radius of gyration as computed. If the 
required area as determined with the actual radius of gyration is less 
than the approximate area, then the thickness of the angles or the 
plates must be increased and the section then examined for its radius 
of gyration and required area. If the area is sufficient, the section 
is used ; if not, another recomputation is in order. 

In the determination of the center of gravity of the section, the 
moment is taken about the top of fhe cover-plate. The moments are 
computed as follows: 

Cover- plate (23 X $) X \ 2 . 88 

Webs 2 (18 X i) X (9 + |) 175.60 

Top angles 2 (2.48) X (1.01 + i) 7.50 

Lower angles 2 (2.48) X (i + i + 18 + \ - - 1.01). . . . 89.30 
Flats" 2 (4 X H) X 19$ 196 . 25 

Total ...... 471.53 

The center of gravity is now found to be i^-^r =9.55 

~ 4 /\ .w.4o 

inches from the top of the cover-plate. The distance from the top of 
the cover-plate to the middle line of the web is 9 + \ -f- % = 9.75 
inches, and this leaves a distance of 9.75' 9.55 = 0.2 inch from 
the center line of the web to the neutral axis. This distance is gen- 



erally represented by the letter e, and it is known as the eccentricity of 
the section. 

The moment of inertia about this axis must now be computed. 
The relation used is that the moment of inertia about any axis is 
eq-ual to the moment of inertia about some other axis, plus the product 
of the square of the distance between the two axes by the area of the 
section whose moment of inertia is desired. The moments of inertia 
of the various parts of the section (see "Steel Construction," Part 
IV, pp. 292 and 293) are computed and are as follows: 

Cover-plates 955 . 26 

Webs 486 . 72 

Top angles 325 . 74 

Lower angles I 359 . 74 

Flats... ..-1017.37 

Total. . .. 3 184.83 

The radius of gyration is equal to the square root of the quotient 
obtained by dividing the moment of inertia by the area. It is 

184.83 =8Q4 

\ 49.42 

Using this value of the radius of gyration in the formula for the com- 
pressive stress, there is obtained 13 800 pounds as the unit allowable 

, 449 500 

stress in compression, and this requires an area of =32.5 

13 800 

square inches. Since this is considerably less than the actual area of 
the section, the section will not be changed but will be taken as first 

In order that the section should be safe about both axes, the 
moment of inertia about the axis perpendicular to the cover-plate 
should be equal to or greater than that as above computed. By com- 
puting the moment of inertia about the axis perpendicular to the 
cover-plate, it is found to be 3 256 . 3, which gives a radius of gyration 
of 8 . 11 ; and since both of these are greater than those first computed, 
it is seen that the section is safer about the axis perpendicular to the 
cover-plate than it is about an axis perpendicular to the web plates. 

There are small stresses in this member due to its own weight 
and to the fact that the pins are not placed directly upon the neutral 
axis (see "Strength of Materials," p. 82). These stresses are seldom 
more than 1 000 pounds per square inch in the extreme fibre; and 


since the section has such an excess of area, they will not be computed, 
as it is evident that there is sufficient strength in the member to with- 
stand them. 

The section just designed is that for the top chord having the 
greatest stress; and since this is the minimum section allowed by 
the Specifications, it must be used in all the sections of the top chord. 

The section as finally designed is: 

One cover plate, 23 by J inch; 
Two webs, 18 by inch; 
Four angles, 3 by 3 by f-inch 
Four flats, 4 by f-inch. 

A pin 6|- inches in diameter will be used at the point U r The 
stress in the member UJJ 2 is 378 200 pounds, and the bearing area 

378 200 
required is - - = 15. 75 square inches, or 7 . 875 for each side. 

7 875 
This makes a total required thickness of ' = 1 . 265 inches for one 

side. Since the thickness of the web plate is | inch, it will be necessary 
to provide pin-plates whose total thickness must be 1.265 0.5 = 
. 765 inch. Two f-inch plates will give a thickness of . 75 inch ; and 
since this is less than the required thickness by an amount not over 
2 .V per cent, they may be used. The total thickness of the bearing 
area is now 1 . 265 inches. The stress transferred to the two f-inch 
plates is: 

X 189 100 = 113 500 pounds. 


The rivets required to keep the outer plate from shearing off the 

113 500 
other are - --- - =8 shop rivets, and the rivets required to keep both 

/\ / Z.kj 

of the -inch plates from shearing off the web of the chord section are 

113 500 

---" =16 shop rivets in single shear. The bearing of a |-inch 

shop rivet on a i-inch plate is 10 500 pounds, and therefore the num- 
ber of rivets required to keep these pin-plates from tearing the rivets 

113 500 

out of the vy-inch web plates is =11 shop rivets in bearing. 

1U olJu 

Fig. 185 shows the detail of this end of the top chord section. The pin- 
plates should extend well back on the member, and at least one pin- 




plate should go over the angle, and enough rivets, as computed above, 
should go through the angles and this pin-plate. Experiments on full- 
sized bridge members go to show that unless the pin-plates cover the 
angles and extend well down on the member, the member will fail 
before the unit-stress reaches that value computed by the formula 
for compression. 

Since the ends of the chord are milled at the splices, and therefore 
butt up against each other and allow the stress to be transmitted 


^O O O O 


u, j 

\ Q. 

| -g Plate 

1 o 


o o 


2 8 Plate'? 

o o 

\| gg 

o o o < 

Fig. 185. Detail of Top Chord Section at Point U v 

directly, only sufficient rivets need be placed in the splice to keep the 
top chord sections in line (55). 

At the point U 2 , it is not necessary to put in a pin-plate to take 
the stress in the upper chord; but it is only necessary to provide a 
pin-plate to take up the difference in stress between the two chord 
sections. This difference in stress is equal to the horizontal com- 
ponent of the maximum stress in the member U 2 L y This is 110 000 
pounds, and the area required on each side for bearing is 2.3 square 
inches; and as a 5-inch pin is used here, the thickness of the bearing 

As this thickness is less than the thick- 

2 3 

area is -^- = 0.46 inch, 

ness of the web plate, no pin-plates will be required. 

At the point U 3 , a bearing area will be required to withstand the 
horizontal component of the member U 3 L 4 . This is 56 300, and the 



bearing area required on each side is 

24 000 X 2 

= 1.18 inches. The 

required thickness of the bearing area is ~ = 0.24 inch, as a 5- 


inch pin is used here also. As this thickness is less than the thickness 
of the web plate, no pin-plate will be required. 

The under parts of these members must be stiffened by tie or 
batten plates, and these plates (44) must be equal in length to the 
distance between rivet lines. This is 1 9 i inches. They will be made 
20 inches long and 23 inches wide. The thickness of these plates (44) 

must be 


0.39 inch (say T V inch). The size of the tie-plates 

will then be 20 in. by T 7 7 in. by 1 ft. 11 in. 

Since the distance between the rivet lines is greater than 15 
inches, double latticing must be used (47); and according to Table 
XXV the lacing must be \ inch 
thick; also, according to (45), it 
must be 2 inches wide, as the 
rivets used are f- inch in diam- 
eter. The lattices will then be 
2-fc by Hn. 

87. The End=Post. Since the 
minimum section as chosen for 
the top chord is about 50 per 
cent in excess of that required by 
the compression formula, it will 


Fig. 186. Calculation of End-Post. 

be assumed to be sufficient for 
the section of the end-post, and 
it will now be investigated to see if it is safe. 

In addition to the stress due to direct compression, the end-post 
is stressed by its own weight, by eccentric loading due to the pin being 
in the center of the web instead of at the center of gravity of the 
section, and to a bending moment at the place where the portal brace 
joins it. This is due to the bending action of the wind on the top 
chord. These different stresses will now be computed; and since the 
post is in all cases stressed by a combination of bending and compres-, 
sive stresses, this fact should be considered in the design. In deter- 
mining the stress in the end-post due to its own weight, the entire 



weight must not be used in computing the bending action, but only 
that component of it which is perpendicular to the end-post. The 
length of the end-post is readily computed, and is as shown in Fig. 
186. The general formula for accurately computing stresses due to 
bending when the member is also subjected to compression, is: 


s = ~rr^T' 

in which, 

S = Stress in pounds per square inch in the extreme upper fibre of the 


M = Exterior moment causing the stress, and is considered positive if it 
bends the beam downward, and negative if it bends the beam up- 

j/,= Distance from the neutral axis to the extreme upper fibre; 

7 = Moment of inertia of the section; 

P = Direct compressive stress, in pounds; 
I = Total length, in inches; 

E = Modulus of elasticity of steel, which is usually taken as 28 000 000 
pounds per square inch. 

In this case the force causing the bending is that component of 
the weight perpendicular to the end-post. This is Wl sin^, in which 
W is the weight of the steel in the end-post; and this is computed and 
is as follows : 

Cover-plate 1 435 Ibs. 

Web plates '. 2 245 " 

Angles 1 250 " 

Flats... ..1245" 

6 175 Ibs. 
Add 25 per cent for details 1 544 ' ' 

Total. ... 7 719 Ibs. 
Substituting in the above formula the various values, there results : 

= j X 7 719 X 36.7 X 0.572 X 12 X 9.55 
, 1Q _ 410500 X (36.7 X 12) 2 

10 X 28 000 000 

= 800 pounds per square inch compression in the upper 
fibre due to bending. 

In the above equation, the stress in the member is 410 500 pounds; 
the distance y l is the distance from the neutral axis to the top of the 
cover-plate, and the coefficient of elasticity of steel is taken as 
28 000 000. 




In computing the stress due to the eccentric loading, the moment 
is equal to the product of the total stress in the member by the dis- 
tance" from the neutral axis to the center of gravity axis causing a 
negative moment. Substituting in the above formula for combined 
stresses, there results: 

- 410500 X 0.2 X 9.55 

S = - 

3 185 - 

410500 X (36.7 X 12)* 

4 70 

10 X 28 000 000 
= 270 pounds per square inch tension in the upper fibre. 

In order to find the compression in the lower fibre, it is only necessary 
to notice that the stresses are proportional to the distances from the 
neutral axis. Accordingly (see Fig. 187), 
the stress in the lower fibre due to the 
weight is 895 pounds tension, and the 
stress in the lower fibre due to the eccen- 
tric loading is 302 pounds compression. 

Before computing the stress due to the 
bending moment caused by the wind on 
the upper chord, it is necessary to in- 
vestigate the post to see if it is fixed or 
hinged at its lower end. This is very 
important, since, if the post is found to 
be hinged, the bending moment will be 
one-half of that which will occur when 
the post is not hinged. 

An end-post is considered hinged when the product of one-half 
of the total stress times the distance between the web plates is greater 
than the product of the wind load acting at the hip, or joint U v times 

the length of the end-post. 

Calculation of Stress 
in Chord. 

T u- i C .. 410500 
In this case the first value is X 

15 = 3075000; and the product of the latter (see Article 29) is 
12 600 X 36.7 X 12 = 5 550 000. Since the latter is greater than 
the former, the post is hinged, and the bending moment at the foot 
of the portal strut, which joins the end-post 28.2 feet from the end, is 
6300 X 28.2 X 12 = 2 130000 pound-inches. The stress in the 
extreme fibre due to this bending moment is: 
2130000 x 11.5 

S = 

3256.3 - 

410500 X (36.7 X 12) a 
10 X 28 000 000 



= 8 250 pounds per square inch tension or compression. 
In computing this stress due to the wind moment, care must be taken 
to take y 1 equal to one-half the width of the cover-plate, and to take 
'the moment of inertia as that about the axis perpendicular to the 

In computing the total stress on the extreme fibre, it must be 
noted that the stresses due to weight and eccentric loading do not 
stress the same extreme fibres as the stress due to wind, the former 
stressing the extreme fibres on the top and bottom of the post, while 
the latter stresses those on the inner and outer sides. The total 

410 500 
direct unit-stress is = 8310 pounds per square inch; and 

this, added to the 8 250 pounds per square inch due to the wind, 
gives a total of 16 560 pounds per square inch on the extreme 
fibre only. 

oc 7 v 1 9 

The allowable unit-stress is 16 000 - 70 X - = 12 200 


pounds per square inch when wind is not taken into account, and 
(23) is H X 12 200 = 15 250 pounds per square inch when the wind 
is taken into account. The difference between this and the actual 
stress is 16560 15250 = 1 310 pounds per square inch, which 
shows that the section is not strong enough. The section can be in- 
creased by widening the cover-plate or by making the plates thicker ; 
but as this excess is due to wind only, the section being amply suffi- 
cient under the other stresses, and is fixed to some extent by the 
floor-beam connection, no change will be made. 

The pin at each end of the end-post will be the same namely, 
6^ inches in diameter and therefore the pin-plates will be the same 
at each end. The total stress in the post is 410 500 pounds, which 

, 410 500 , 

makes a required bearing area of = 17.2 square inches for 

^4 UUU 

both sides, or 8.6 square inches for one side, and the total required 

8 6 
thickness of ' = 1.375 square inches for one side. Since the 

thickness of the web plates is \ inch, this leaves a remainder 
of 1.375 0.5 = 0.875 inch for the thickness of the pin-plates. 
One plate f inch thick and one plate \ inch thick will be used. 

The proportion of the total stress which is taken by the f-inch 




plate is X = 56 000 pounds; and that taken by the 

1 . o75 2i 

-inch plate is ^ 7 ~X 205250 - '74 600 pounds. The number of 

rivets required to transfer the stress from the f -inch plate to the ^-inch 

56 000 
plate is = 8 shop rivets in single shear; and the number of 

rivets required to transfer the stress from both pin-plates to the web is 

56 000 + 74 600 , L 

=18 shop rivets in single shear. As in the case of 

the top chord, one pin-plate should extend over the angle, and the 
number of rivets required in that pin-plate should go through the pin- 
plate and the angles (see Fig. 
188). The |-in. hinge plate is 
used for erection purposes, and is 
not considered as a pin-plate. It 
is omitted at L . 

Since this section is the same 
as that of the top chord, the tie- 
plates and the lattice bars must 
be the same size. 

88. The Pins. The design of 
the pins requires a simple but 
quite lengthy computation. Sim- 
ple Pratt railroad trusses for 
single-track bridges usually have 
the same arrangement of tension 

and compression members; that is, the same tension members occupy 
relatively the same positions with respect to the compression mem- 
bers. Also,while theoretically a different sized pin will be required at 
every joint, it is not customary to make them so. In practice the 
pins at the joints U l and Z/ are made of the same diameter, and 
those at the remainder of the joints are also made in diameter equal 
to each other but different from those at U 1 and L , the pins at U\ 
and L usually being larger in diameter. On account of the above 
conditions and facts, it is unnecessary to design the pins in spans 
under 200 feet, since usually they are the same for any given span 
and loading. Table XXVI gives the diameters of pins for spans of 
100 up to 200 feet for loading E 50. 


Plates and Riveting at Upper 
End of End-Post, 




Pins for SingIe=Track Bridges 

Loading E 50 


U 1 and L 

All Others 

100 feet 

4i inches 

4 inches 

125 " 

5* " 


150 " 

6J " 

5i " 

175 " 

6| " 

52 ' 

200 " 



For E 40 loading, decrease the above values by J inch; for E 30 loading, 
decrease them by f inch. The diameter of pins for spans not given in the 
table can be interpolated from the given values. No pin should be less than 
3 inches in diameter. 

The span of this bridge is 147 (say 150) feet, and the diameter 
of the pins at U l and Z/ is 6^ | = 6J inches; and the diameter 
of the pins at the other panel points is 5| J = 5 inches. It 
should be noted that no pin is required at point L v as the two mem- 
bers which join here are built-up members and are riveted together. 

The above table is for single-track bridges only. The diameters 
of pins for double-track bridges are given in Table XXVII. These 
values are for E 50 loading ; and for E 40 and E 30 loading, deduc- 
tions must be made as required in the case of Table XXVI. 

Pins for Double-Track Bridges 

Loading E 50 


E7, and L 

All Others 

100 feet 

6 inches 

5} inches 

125 " 


6j " 

150 " 


1\ " 

175 " 

9} " 

81 " 

200 " 

9i " 


No pin in a double-track bridge should be less than 4 inches in diameter. 

Pins for highway bridges are usually much less in diameter than 
those for railway bridges, except in the case of first-class trusses for 
heavy interurban traffic or for city bridges carrying paved streets. 




where they should be taken equal to those given for E 30 loading. 
Table XXVIII gives the diameters of pins for different length spans 
of simple highway bridges designed for 16- ton road-rollers or farm 
wagons and 100 pounds per square foot of roadway. 

Pins for Country Highway Bridges 


U , and Lower Chord 

Upper Chord 

50 feet 

2 inches 

2 inches 

100 " 


2i " 

150 " 

3i " 

2| " 

200 " 


3 " 

89. The Portal. In order to have a clearance of 21 feet (2) 
above the top of rail, it is necessary that the portal be as shown in 
Fig. 189. The stresses are found 
by methods of Article 54, Part I, 
the wind load being computed 
according to (10). It must be 
remembered that the column is 

In case the members of the 
portal braces bend about one 
axis, their length will be equal 
to the distance from one end to 
the other. In case they bend 
about the other axis as indicated 
by the broken line in Fig. 189, 
their length will be one-half of 
what it was in the first case. 

The portal struts or 
diagonals will be designed first. 
Their length is 8.5X1.414- 
12 feet, or 144 inches. This is 

Fig. 189. Portal Dimension and Stress 

the total length. Although 
the Specifications do not men- 
tion it, the ratio of the length to the radius of gyration should 
not exceed 120. This means that the radius of gyration in this 



case should be greater than- = 1.2. The section of the strut will 

be composed of two angles placed back to back. 

Two angles 3J by 3 by f-inch, with an area of 4.6 square inches 
and r 2 equal to 1 .72 see Carnegie Handbook, p. 146, and (72) will 
be assumed to be sufficient to take the stress, and they must now be 
examined to see if the assumption is correct. 

The allowable unit-stress (23) is 25 per cent greater than in the 
case of live or dead loads. This makes the unit-stress as computed 
from the formula: 

16 000 - 70 x 

j \\ = 12 680 pounds per square inch. 

*3S 500 
The required area is ' = 3.05 square inches; and since this is 

less than the given area, the angle will be amply sufficient. The re- 
quired area is over one square inch less than the given area, but this 
angle must be used, since it is the smallest angle allowed by the Specifi- 
cations. Note that unequal legged angles should be used, as this will 
make the radius of gyration about one axis larger than about the other; 
and this will prove economical, since, when one axis is considered, 
the length of the member is greater than when the other is considered. 
The above angle should also be examined for tension, it being 
considered that one rivet-hole is taken out of the section of each angle. 
The net section of the two angles will now be 4.60 - 2 (| + ) X f 

38 500 
= 3 . 85 square inches ; and the area required for tension is ftnm 

= 1 .93 square inches, which shows that the angle is amply sufficient. 
It should be noted that these Specifications do not require that only 
one leg of the angle shall be efficient unless both legs are connected. In 
case this strut had been designed according to Cooper's Specifica- 
tions, two angles 5 by 3 by ^-inch would have been required, and the 
5-inch leg would have been placed vertically and the angle connected 
by this leg alone. While it is not within the province of this work to 
discuss the question of connecting angles by one or by both legs, yet 
it might be said that tests made on angles connected with one leg 
only, seem to indicate that the ultimate strength in tension is about 
60 per cent of that obtained from the same angle when tested with 
both legs connected. 



While according to (20) the alternate strains in the wind bracing 
do not have to be considered, since they do not occur very closely 
together, yet in framing the connections it is required that the sum of 
both positive and negative stresses shall be added. In this case the 
stress for which the connections must be designed is 2 X 38 500 
= 77 000. It must be remembered that in this case also, the 
unit-stresses are increased 25 per cent over those allowed for live and 
dead loads. 

The number of rivets required in the end connections will be 
governed by bearing in the connection plates, and these plates are 
usually made f-inch thick. The number of rivets required is 

77 000 77 000 

TSSO^Oi = 8 Sh p nV6tS ' r 6560XT* = 

The portal bracing is riveted up in the shop and brought to the 
bridge site, where it is connected to the trusses by field-riveted con- 
nections at its end. Therefore the end of the portal struts which 
connect with the top piece will have 8 shop rivets, and the other end 
which connects with the end-post will have 10 field rivets. Since 
the angles are small, all the above rivets must go in one line, and this 
will cause the connection plate to be quite large. It will probably be 
better to connect both legs of the angle by means of clip angles and 
thus reduce the size of the connection plates. 

The top part of the portal bracing will consist of two. angles. 
Two angles 3^V by 3 by f-inch will be assumed and examined to 
determine if the area is sufficient. The length of this strut is the 
distance center to center of trusses, and is equal to 17 X 12 = 204 

inches. The least radius of gyration is therefore *"- = 1 . 70. The 

radius of gyration of the two angles assumed is 1 . 72 when referred to 
an axis parallel to the shorter leg when the two angles are placed back 
to back and One-half inch apart. The unit-stress is now computed : 

P = Ae 000 - 70 X ^2) H = 9 625 pounds per square inch. 

27 200 

The required area is " = 2.825 square inches. This is con- 

siderably less than the area given by the two angles; but as these are 
the minimum angles allowable, they must be used. Since the stress 
in this case is less than in the previous -case, and since the angles 




used are the same, it is evident that these angles are safe in tension. 
The number of rivets is determined by the bearing in the f-inch 
connection plates, and is : 

2 X 27 200 

7880 X 1.25 
2 X 27 200 
6 560 X 1 . 25 

6 shop rivets, and 

= 10 field rivets. 

As in the case of the lateral strut, this member should be connected 
by both legs of the angle in order to reduce the size of the connection 

plates. Fig. 190 gives the 
details of the portal bra- 
cing and its method of 
connection to the end- 
post. The full circles 
represent shop rivets, and 
the blackened circles rep- 
resent field rivets. Some 
engineers connect the 
portal bracing to the top 
cover-plate of the end- 
post. This produces an 
excessive eccentricity in 

Fig. 190. Details of Portal Bracing and Connection the end-DOSt and is bad 
to End-Post. 


Those members of the portal bracing which do not take any 
stress will be made of single angles, and the size of these angles will be 
taken 3 by 3 by f-inch. 

90. The Transverse Bracing. This bracing will be the same 
general style as the portal bracing, except that the top member will 
consist of two angles placed at a distance apart equal to the depth 
of the top chord, and these angles will be joined together by lacing. 
As in the case of portal bracing, those members which do not take 
stress will be made of one angle 3| by 3 by f-inch. 

The general outline is shown in Fig. 191, and the stresses are com- 
puted from (10) and by the methods of Article 54, Part I. In design- 
ing this top member, the top angle only is supposed to take the stress. 
The length in this case is 204 inches. Two 3% by 3 by f-inch angles 
will be assumed as sufficient and will be examined. These angles 




give a total area of 4.60 square inches. In examining these it will 
be found that they are amply sufficient, in fact so much so that it will 
be better to see if one single angle at the top will not be better. 
According to the length, the smallest radius of gyration which can be 
used is 1.7. In looking over the tables of angles, it is seen that the 




Section ot BW 

Fig. 191. General Outline of Transverse Bracing. 

first angle to fulfil this condition is a 6 by 3^ by f-inch, and it has a 
radius of gyration of 1.94. The allowable unit-stress is computed 
as follows: 

6 000 - 70 X 

= 10 78 pounds per square jnch; 

and the required area is -=~; = 0.85 square inch. This is con- 




siderably smaller than the area of the angle, which is 3.97 square 
inches; but since this is the smallest possible angle which will fulfil 
the conditions of the Specifications, and since it is much smaller than 
the two angles as first assumed, it will be used. Fig. 192 gives a cross- 
section of this member. Since this angle is 
joined to the cover-plate by one leg, the joints 
will be weak in single shear, and the number of 
rivets required will be: 

2 X 9 100 

2 X 9 100 
6013 X H 

3 field rivets. 

According to (45), the width of the latticing 
must be 1\ inches; and according to Table XXV, 
the thickness must be -, 7 B inch, the distance c be- 
ing 1 foot 11 inches. 

The length of the knee-bracing is 144 inches; 
but on account of the small stress, one angle will 
be used. One 4 by 3 by 4-inch angle, with an 

*, ' * 

area of 2 . 48 square inches and a radius of gyra- 
tion 1.26, will be assumed as sufficient. The radius of gyration 
is greater than the minimum allowable, which is 1.2. The allowable 
unit-stress is: 

Pig. 192. Cross-Sec- 

tion of Top Member or 

Transverse Bracing. 

P = ( 1C 000 - 70 X y-Q?;) H = 10000 pounds per square 


The required area is 


= 1 . 23 square inches. The required 


area is much less than the given area; but this angle must be used, 
since it is the only one allowed on account of its radius of gyration. 
Two of the minimum sized angles might have been used; but their 
total area, 4.60 square inches, is much in excess of that of the angle 

This angle must be examined for tension. The net area is 2.48 
- (I + i) X I = 2 . 1 square inches. The required net area in 

12 300 

tension is ~ L = 0.615 square inch, which shows this angle 

ID uuu /\ i j 

to be amply sufficient. 




The number of rivets required will be governed by the shear, 
since the angle is connected by one leg only; and it is: 

2 X 12 300 

7 220 X H 
2 X 12 300 
6013 X H 

= 3 shop rivets, and 

= 4 field rivets. 

91. The Lateral Systems. The stresses in these systems must 
be computed according to (10) and Article 54, Part I. They are given 
on the stress sheet, Plate III (p. 251). Since according to (68) these 
members must be constructed of rigid shapes, it is customary, in com- 
puting the stresses, to assume that one-half the shear is taken by each 
of the diagonals in any given panel; that is, one diagonal is in tension, 
and the other diagonal is in compression. The 
stresses given on the stress sheet are computed 
by making this assumption. Also, since both 
diagonals in each panel are considered as acting 
at the same time, the stresses in all the verticals 
are zero. 

The section of the upper lateral members will 
be made up of two angles placed apart a distance 
equal to the depth of the top chord. Fig. 193 
shows the section. The radius of gyration about 
the axis parallel to the long leg will be consider- 
ably larger than that about an axis parallel to the 
shorter leg. In fact, it is so much greater that the 
strut will not need to be examined with respect 
to this axis. The diagram of the first panel is 
given in Fig. 194. The radius of gyration is to be taken about 
the horizontal axis if the entire length is to be taken; and the radius 
of gyration is to be taken about the vertical axis if one-half the 
length is taken, in which case it will bend as shown by the broken 
line in Fig. 194. The members are designed for the latter 
conditions only, since they are amply safe in regard to the first 
condition if they satisfy the latter. The length in this latter con- 
dition is 13.5 feet, w r hich requires a radius of gyration not less than 
13.5 X 12 


Section of 

Upper Lateral 



Two angles 5 by 3 by f-inch, with a total area of 5 . 72 square 




inches and a radius of gyration equal to 1.61, will be assumed and 
investigated to determine if they are sufficient. 

1Q (% \x 19 

The unit-stress is computed to be P = (16 000- 70 X p^") X 

A 7OO 

1 \ = 11 000 pounds per square inch, and the required area is = 

1 1 UUU 

0.61 square inch. The required area is very much less than the 
given area; but the angle chosen must be used, since this is the 
smallest one which conforms to the requirements of the Specifications. 

Fig. 194. Outline Diagram of First Panel 
in Upper Lateral System. 


Fig. 195. Outline Diagram of First Panel 
in Lower Lateral System. 

The width of the lattices (46) must be 2| inches; and according 
to Table XXV, the thickness must be -^ inch, the distance c being 
23 inches. 

Single shear governs the number of rivets required. In accord- 

ance with (20) and (23), their number is 

= 2 rivets. Field 

6 013 X 1 j 

rivets 3 in number are used in all places, since the lateral system is 
riveted up after the trusses are swung into place. 

Since this is the minimum sized angle which will give a radius of 
gyration greater than 1 . 35, it must be used in the remainder of the 
panels of the top chord. Four angles of the minimum size might 
have been used, and would have been satisfactory, except that the 
area would have been excessive. 

The stresses in the lower lateral system are computed according 
to (10), a similar assumption to that for the upper lateral system being 
made namely, that both diagonals in each panel are stressed at the 
same time, one taking tension and the other taking compression. 
Fig. 195 shows the first panel of the lower lateral system. These 



diagonals are connected to the stringers wherever they cross them, 
and also to each other where they cross in the center. This reduces 
the length which must be used in computing the cross-section of the 
member. In this case it is the distance C-A, and is equal to 90 inches. 
Since the angle is free to move about either axis, angles with even 
legs should preferably be employed, since this will give greater 

economy. The radius of gyration must be greater than - = 0.75. 

One angle 3 1 by 3^ by f-inch, with an area of 2 .48 square inches 
and a radius of gyration of 1.07, will be assumed and investigated. 

The allowable unit-stress is P = (16 000 - 70 X ) 1| = 12650 

30 500 
pounds per square inch, and the required area is = 2 . 38 square 

inches. This is nearly equal to the given area, and therefore the 
angle chosen will be taken for the section. 

This angle must now be investigated for tension, one rivet-hole 
being taken out of the section. The net area is 2.48 (J + i) f = 

on Af) 

2.10 square inches. The required net area is ^ -- - =1.53 

square inches, which shows the angle to be sufficiently strong. 

Single shear determines the number of rivets to be required. 
These are : 

All rivets in the lower lateral system are field rivets, since this system 
also must be riveted up in the field after the trusses are swung into 

The total stress in the second panel is 21 500 pounds, and a 
3 by 3 by f-inch angle, with an area of 2.30 square inches and a 
least radius of gyration of . 90, will be assumed and examined. The 

allowable unit-stress in compression is 1^ (16 000 70 X ) = 

21 500 
1 1 250 pounds per square inch, and the required area is = 1 .91 

square inches. Since this is less than the given area, and since the 
size of the angle (72) is the smallest allowable, this angle must 
be used. 




It is required that this member shall have a net area of 
91 c^oo 
- = 1.08 square inches in tension. The net area of the 

angle, one rivet-hole being taken out, is 1.92 square inches, which 
shows the angle to be safe in tension. 

The number of rivets required is determined by single shear, 

Fig. 196. Two Types of Bearings. 

since they tend to shear off between the member itself and the con- 

2 X 21 500 

necting plates. The number required is 77:^ = 6 field rivets. 

6 Olo X lj 

Since the above angle is the smallest that can be used, and since 
the remaining angles of the panel of the lateral bracing have smaller 
stresses than the one just designed, it is evident that this size angle 
must be used in all panels of the lower lateral system other than the 

92. The Shoes and Roller Nests. For bridges of short spans 
and for plate-girders whose spans require rocker bearings to be pro- 
vided (80), several different classes of bearings are in use. Two such 
bearings are shown in Fig. 196 (a and b). The type illustrated by a 
is seldom used on any spans except plate-girders. That shown in b 



may be used on either plate-girders or small truss spans; it is the 
invention of Mr. F. E. Schall, Bridge Engineer of the Lehigh Valley 
Railroad, who uses it on plate-girders. It has given very great satis- 
faction ; and for simplicity of design and also for economy it is to be 
recommended. Some railroads have used a bearing which consisted 
of a lens-shaped disc of phosphor-bronze, the faces of which fitted 
into corresponding indentations in both the masonry and the bearing 
plates. One advantage of this bearing is that it allows movements 
due to the deflection of the girder, and also lateral deflection of the 
floor-beam. It is claimed to have given satisfaction. 

A bearing which is used on both short-span and long-span bridges 

Fig. 197. Bearing Adapted to Bridges of Both Short and Long Span. 

is shown in Fig. 197. This class of bearing will be used. The end 
reaction of the bridge proper is equal to the vertical component of the 

30 1 
stress in the end-post, and is -^ - X 410 500 = 336 500 pounds, 

336 500 
which requires a bearing area (19) on the masonry, ot ^- =561 


square inches. According to the table on page 193, the masonry 
plate will be 28 inches long. 

The total bearing area for one of the vertical plates is: 



and the total required thickness is: 

1^=1.12 inches, 

a 6^-inch pin being used at L . Since the vertical plates will be made 
f inch thick, this leaves a remainder of | inch to be made up of pin- 

The amount of stress which is carried by the f-inch pin-plate is 

= 56 100 pounds. These plates will tend to shear 

56 100 
off the rivets at a plane between the plates, and therefore = 8 

shop rivets will be required to fasten them to the vertical plate. 

Since the length of the masonry plate is 28 inches, and the total 

area required is 561 square inches, the required width is = 20 

inches. The actual width will be greater than this, since it must be 
sufficient to allow for the connecting angles and also for the bearings 
of the end floor-beam. The connecting angles should be f inch thick, 
and should not be less than 6 by 6 inches; and the plates to which they 
are connected should not be less than f inch in thickness, and likewise 
they should not be greater, on account of the punching. The bottom 
plate should extend outward about 3 inches, in order to allow suf- 
ficient room for the anchor bolts, which should be f inch in diameter 
and should extend into the masonry at least 8 inches. 

In addition to the reaction of the bridge proper, the masonry 
plate must be of sufficient area to give bearing for the end reaction of 
the end floor-beam. The maximum end reaction (see Article 83, p. 
197) is 104 740 pounds. The bearing area required on the masonry 

104 740 

is = 175 square inches; and assuming that the base of the 

bearing will be 12 inches long (see Fig. 197), the required length will 
be 14.6 inches. Usually, however, the bearing is extended the entire 
length of the masonry plate, which is 28 inches in this case. 

The distance from the center of the pin to the top of the masonry 
will be the same for both the fixed and the roller end. This distance 
should be such that the angles of the shoe will clear the bottom chord 
member and allow the floor-beam to rest upon the plate as shown. 
Since the first section of the bottom chord is 18A inches deep, the top 
of the angles of the two must be at least 9j inches from the center line 




of pins. This requires that the distance from the center line of the 
pin to the base of the angle shall be at least (9j + 6) = 15j inches, 
or more. 

The tops of all floor-beams are at the same height, and the 
bottoms of the intermediate floor-beams must be on a level with the 
bottom of the first section of the lower chord (see Fig. 174). This 
requires that the bottom of the intermediate floor-beams shall be 
9j inches below the center line of pins, and this brings the top of the 
floor-beams (48-f 9|) = 39 inches above the center line of the pins. 
Since the end floor-beam is 52 \ inches deep, back to back of angles, the 

Fig. 198. Type of Bearing Construction where End Floor-Beam Does Not Rest Directly on 
Bearing or Masonry Plate. Grillage of Iron Bars Used instead of Cast-Steel Pedestal. 

lower flange will be (52 J 39) = 13| inches below the center line of 
pins. In case the end floor-beam does not rest directly upon the 
bearing plate or the masonry plate, the intervening space is filled out 
with a grillage of iron bars or a cast-steel pedestal, as shown in Figs. 
197 and 198. 

The small plates upon the side of the shoe, going entirely around 
the pin, are called the shoe hinge-plates. These do not take any stress, 
and require only sufficient rivets to hold them in position. They are 
used during erection to keep the end-post in line; and after erection 
their function is to keep the end-post on the shoe, and to prevent it 
from having any upward motion due to the vibration of the structure. 



The rivets through the vertical legs of the shoe angles are in 
double bearing in the f-inch angles, in single bearing in the vertical 
plate, and in double shear. A rivet in double shear has a less value 
than in bearing in the plates. This value is 14440 pounds, and 
therefore the number of shop rivets required through the vertical legs 
of the angles is : 

336 500 
2^04440 =12nVetS - 

The rivets which go through the horizontal leg of the angle and 
through the cap plate and cap angles, do not take stress. The num- 
ber of rivets put in is that demanded by the detailing, the rivets in the 
horizontal legs of the angles usually staggering with those in the verti- 
cal legs. The cap plate tends to keep the vertical plates in line, and 
to keep out the dust and dirt and other deteriorating influences of 
the elements. 

Wherever the rivet-heads tend to interfere with other members 
or project beyond surfaces which are required to be flat as, for 
example, the bottom of the masonry or bearing plates they must be 
countersunk (see Carnegie Handbook, p. 191, and "Steel Construc- 
tion," Part III, p. 192). 

The space for the anchor bolts, that for the connection angles, 
and that for the bearing of the end floor-beam, require that the total 
width of the masonry plate for the fixed end shall be 2 X f + 14 + 2 
X 6 + + 3 + 12 - 3 feet 7$ inches. 

The design of the roller end requires that the length of the 
masonry bearing, the size of the vertical plates and angles, and also 
the number of rivets shall be the same as that for the fixed end. The 
width of the masonry plate is determined by the length of the rollers 
and their connections at the end. 

The rollers (60) are required to be 6 inches in diameter, and the 
unit-stress (19) per linear inch is 6 X 600 - 3 600 pounds, which 
requires : 

336 500 

n-fnfr = 9- linear inches, 
o oUU 

This is for the reaction of the bridge alone; and in addition to this, 
there are required for the floor-beam reaction : 
104 740 


= 29.0 linear inches. 




The total number of linear inches is 93.5 + 29.0 = 122. 5; and if 5 

122 5 

rollers are used, they must be at least ^- = 24.5 inches long. The 


masonry plate is only 28 inches long, and therefore cylindrical rollers 
cannot be used, since they would occupy a space 30 inches or over. 
Segmental rollers (see Fig. 199) 
must be used. 

The determination of the 
sizes of the angles which go at the 
end of the rollers, and also of the 
guide-plates, is a matter of judg- 
ment and experience. Those 
sizes indicated in Fig. 198, repre- 
sent good engineering practice, 
and will be used. 

The distance from the center 
line of pins to the top of the ma- 
sonry can now be determined, 
and is 1G| + f -f 6 + f = 23f 

On account of putting in Fig. 199. Segmental Rollers Used for Bear- 
ings in Space under 30 Inches. 

sufficient connections and angles 

as shown in Fig. 198, the masonry plate must be considerably wider 
than that theoretically determined. According to Fig. 198, the total 
width must be as follows, and the width should be computed in two 
parts, as the plate is not symmetrical about the center line of the truss: 
From center line to outer edge: 

^ + I + 6 + + (3 - $ = 2f ) + (3$ - | = 3i) + 3 

= 1 ft. Hi in., (say, 1 ft. 11 in.). 
From center line to inner edge: 

+ | + 6 + 1 + 12 + + 2$ + 3i + 3 = 3 ft. Oi in., say, 3 ft. in. 

Total width 4 ft. 11 in. 

Allowing guide-plates and guide-bars of dimensions as shown in 
Fig. 198, and assuming | inch as clearance at the ends, the total length 
of the rollers is: 

(4ft. 11 in. ) - 2(3 + 3i + | + | + J) = 44. 5 inches. 
This shows them to be amply long enough, as only 22 inches is 



theoretically required. Here, as in most cases for single-track spans 
up to 200 feet in length, the width of the masonry plate is determined 
by the detail and not by the unit bearing stress. 

The guide-plates are small bars riveted to the top of the bottom 
plates, and serve to keep the rollers in line. The guide-bars are con- 
nected to rollers at their ends, and serve to keep the rollers equi- 
distant, therefore causing them to roll easier and keeping them from 
becoming worn by contact with each other. 

The expansion (57) must be allowed for at the rate of | inch for 
every 10 feet in length of span. This makes a total allowed for tem- 

perature of expansion of --_- X & = If (say 2) inches. No slotted 

holes are to be provided for the anchor bolts, since they do not go 


i i i ! ' 


. M (-7 
<.: "";!-"*-,.' I 


Fig. 200. Binding of Insufficiently Spaced Fig. 201. Computation of Spacing for Seg- 
Segmental Rollers. mental Rollers. 

through that part of the bridge which slides. The shoe slides over 
the rollers, and is kept in place by the angles at the end, which are 
riveted to the masonry plate (see Fig. 198). 

Unless sufficient room is allowed between the segmental rollers, 
they will tend to bind when the bridge has reached the extreme posi- 
tion for expansion or contraction (see Fig. 200). This distance can be 
computed from proportions as indicated in Fig. 201, and from the 
following formulae*: 

in which e is the amount allowed for the change of temperature, and 
D is the diameter of the rollers, both being taken in inches. The 
angle <f> is in degrees. In the present case, e'is2 inches; D is 6 inches; 

* Derived by Prof. Frank B. McKibben of Lehigh University, and published in 
Engineering Xews, December, 1896. 



and <, computed from the above formula, is 9 30'. Substituting in 
the equation giving the value for y, there is obtained 1.02 inches (say 
\\ inches) for the distance between rollers. Rollers must not be less 
in thickness than the total expansion allowed for temperature. 

Since there are 5 rollers, there are 4 spaces between them. Also, 
since the rollers must occupy a space of 28 inches, the length of the 
masonry plate, each roller must be: 

OQ _ 4 v 1 4- 

-1=4.6 inches (say 4 inches) wide, 

The width of the guide-bars must be such as to allow freedom 

Fig. 202. Details of End Floor-Beam Connections. 

of motion for the rollers. The maximum width allowable is given by 
the formula :* 

W = -^-cos <, 

in which < and D are indicated above. This requires the bar to be . 

W =^~X 0.985 = 2.96 (say 2^) inches wide. 

93. The Stress Sheet. Plate III shows the stress sheet of the 
bridge which has been designed in the preceding articles. This 
sheet represents the best current practice among the larger bridge 

* Derived by Prof . Frank B. McKibben of Lehigh University, and published in 
Engineering News, December, 1896. 



corporations. It will 
be noted that very 
few details are given 
upon the sheet; also 
that few rivets are 
noted, and that 
sketches showing the 
manner in which the 
parts go together are 
entirely wanting. The 
shears and moments 
for the stringers and 
floor-beams, as well as 
the reactions and the 
number of rollers re- 
quired, are given. This 
is to save the drafts- 
man the trouble of 
recomputing values 
which have necessarily 
been determined by 
the designer. 

The details of the 
various members, and 
also the manner in 
which the different 
members are con- 
nected, are left to the 
draftsman, who is un- 
der the direct super- 
vision of the engineer 
in charge of the draft- 
ing room, upon whom 
rests the responsibility 
for good details. The 
figures given in the 
text indicate the best 
current practice. Figs. 

n _j i 10 

-1 " -1 a 








. Details of 



202 to 204 show details of the end floor-beam connections, and also 
the packing of the members of the upper and the lower chord. The 
arrangement here given may be said to be standard for single-track 
Pratt truss spans up to 200 feet in length. 


The following books are recommended to the student in case it 
is desired to pursue further the study of the subjects of Bridge 
Analysis and Bridge Design : 

The Theory and Practice of Modern Frame Structures. JOHNSON; 
BRYAN, and TURNEAURE. John Wiley & Sons, New York, N. Y. 

Roofs and Bridges. MERRIMAN and JACOB Y. John Wiley & Sons, New 
York, N. Y. 

Design and Construction of Metallic Bridges. BURR and FALK. John 
Wiley & Sons, New York, N. Y. 

Influence Lines for Bridges and Roofs. BURR and FALK. John Wiley 
& Sons, New York, N. Y. 

Details of Bridge Construction. FRANK W. SKINNER. McGraw Pub- 
lishing Company, New York, N. Y. 

Steel Mill Buildings. Milo S. Ketchum. Engineering News Publish- 
ing Company, New York, N. Y. 

Statically Indeterminate Stresses. HIROI. Engineering News Pub- 
lishing Company, New York, N. Y. 

Stresses in Frame Structures. A. JAY DuBois. John Wiley & Sons, 
New York, N. Y. 

Die Zusatzkrdfte und Nebenspannungen eiscrner Fachwerkbrucken. FR. 
ENGESSER. Julius Springer, Berlin, Germany. 

Bridge Drafting. WRIGHT and WING. Engineering News Publishing 
Company, New York, N. Y. 

It must not be presumed that the above is a complete list of the 
books which have been published relating to the theory and practice 
of Bridge Engineering; neither must it be presumed that the obtaining 
of information relative to bridges is limited to textbooks on the 
subject. One of the best sources of information is found in the cur- 
rent engineering periodicals and the "Proceedings" of the various 
technical societies. The great advantage of these sources is that they 
give the most up-to-date information, and usually they are very pro- 
fusely illustrated. 



Section of 35-mile road built by American engineers, connecting the seaport of Dagupan with 
the mountain village of Baguio. province of Beuguet. island of Luz6n, and affording a cool and 
healthful retreat from the heat and malaria of the lowland repions. Dagupan lies 120 miles north or 
Manila, with which it is connected by rail. This view reveals some of the engineering dlfflcoluea to 
be overcome, masonry and concrete work of the type shown being necessary at many points. 




2 2 

8 = 


w 2 

H g 


O < 

O x 







Object of Roads. The object of a road is to provide a way 
for the transportation of persons and goods from one place to another 
with the least expenditure of power and expense. The facility with 
which this traffic or transportation may be conducted over any given 
road depends upon the resistance offered to the movement of vehicles. 
This resistance is composed of: (1) The resistance offered by the 
roadway, which consists of (a) "friction" between the surface of 
the road and the wheel tires; (6) resistance offered to the rolling of 
the wheels, occasioned by the want of uniformity in the road surface, 
or lack of strength to resist the penetrating efforts of loaded wheels, 
thus requiring the load to be lifted over projecting points and out of 
hollows and ruts, thereby diminishing the effective load the horse 
may draw to such as it can lift. This resistance is called "resistance 
to rolling" or "penetration;" (c) resistance due to gravity called 
"grade resistance;" (2) The resistance offered by vehicles, termed 
"axle friction;" (3) Resistance of the air. 

The road which offers the least resistance to traffic should com- 
bine a surface on which the friction of the wheels is reduced to the 
least possible amount, while offering a good foothold for horses, to 
enable them to exert their utmost tractive power, and should be so 
located as to give the most direct route with the least gradients. 

Friction. The resistance of friction arises from the rubbing of 
the wheel tires against the surface of the road. This resistance to 
traction is variable, and can be determined only by experiment. 
From many experiments the following deductions are drawn : 

(1) The resistance to traction is directly proportional to the 

Copyright, 1!>08, by American School of 



(2) On solid, unyielding surfaces it is independent of the width 
of the tire, but on compressible surfaces the resistance decreases as 
the width of the tire increases (but there is no material advantage 
gained in making a tire more than 4 inches wide). 

(3) It is independent of the speed. 

(4) On rough, irregular surfaces, which give rise to constant 
concussion, it increases with the speed. 

The following table shows the relative resistance to traction of 
various surfaces: 

Resistance to Traction on Different Road Surfaces. 

Pounds per ton. 

In terms of load. 

Earth road ordinary condition 

50 to 200 
50 to 100 

A to T V 
:nr to ? V 


100 to 200 

1 tO T 1 * 


30 to 100 

5t to 

Plank Road 

30 to 50 

JL to A 

Steel Wheelway 

15 to 40 

i to A 

Traction Resistance. 

These coefficients refer to the power required to keep the load 
in motion. It requires from two to six or eight times as much force 
to start a load as it does to keep it in motion, at two or three miles 
per hour. The extra force required to start a load is due in part 
to the fact that during the stop the wheel may settle into the road 
surface, in part to the fact that the axle friction at starting is greater 
than after motion has begun, and further in part to the fact that 
energy is consumed in accelerating the load. 

Resistance to Rolling. This resistance is caused (1) by the 
wheel penetrating or sinking below the surface of the road, leaving 
a track or rut behind it. It is equal to the product of the load mul- 
tiplied by one-third of the semi-chord of the submerged arc of the 
wheel; and (2) by the wheel striking or colliding with loose or pro- 
jecting stones, which give a sudden check to the horses, depending 
upon the height of the obstacle, the momentum destroyed being 
oftentimes considerable. 

The rolling resistance varies inversely as some function of the 



diameter of the wheel, as the larger the wheel the less force required 
to lift it over the obstruction or to roll it up the inclination due to the 
indentation of the surface. 

The power required to draw a wheel over a stone or any ob- 
stacle, such as S in Fig. 1, may be thus calculated. Let P represent 
the power sought, or that which would just balance the weight on 

the point of the stone, and the 
slightest increase of which 
would draw it over. This 
power acts in the direction 
C P with the leverage of B C 
or D E. Gravity, represented 
by W, resists in the direction 
C B with the leverage B D. 
The equation of equilibrium 

Fig. 1. 

will be P X C B = W X B D, whence 



I^et the radius of the wheel = C D = 26 inches, and the height 
of the obstacle = A B = 4 inches. Let the weight W = 500 pounds, 
of which 200 pounds may be the weight of the wheel and 300 pounds 
the load on the axle. The formula then becomes 


26- 4 



314.7 pounds. 

The pressure at the point D is compounded of the weight and 
the power, and equals 



500 X 

591 pounds, 

and therefore acts with this great effect to destroy the road in its 
collision with the stone, in addition there is to be considered the 
effect of the blow given by the wheel in descending from it. For 
minute accuracy the non-horizontal direction of the draught and 
the thickness of the axle should be taken into account. The power 
required is lessened by proper springs to vehicles, by enlarged wheels, 
and by making the line of draught ascending. 


The mechanical advantage of the wheel in surmounting an 
obstacle may be computed from the principle of the lever. 

Let the wheel, Fig. 2, touch the horizontal line of traction in 
the point A and meet a protuberance B D. Suppose the line of 
draught C P to be parallel to A B. Join C D and draw the perpen- 
diculars DE and D F. We 
may suppose the power to be 
applied at E and the weight at 
F, and the action is then the 
same as the bent lever E D F 
turning round the fulcrum at 
D. HenceP:W::FD :DE. 
ButFD :DE ::tanFCD:l, 
and tan F C D = tan 2 
(DAB); therefore P - W 
tan 2 (DAB). Now it is obvious that the angle DAB increases 
as the radius of the circle diminishes; and therefore, the weight W 
being constant, the power required to overcome an obstacle of given 
height is diminished when the diameter is increased. Large wheels 
are therefore the best adapted for surmounting inequalities of the 

There are, however, circumstances which provide limits to the 
height of the wheels of vehicles. If the radius AC exceeds the 
height of that part of the horse to which the traces are attached, 
the line of traction C P will be inclined to the horse, and part of the 
power will be exerted in pressing the wheel against the ground. The 
best average size of wheels is considered to be about 6 feet in diameter. 
Wheels of large diameter do less damage to a road than small 
ones, and cause less draught for the horses. 

With the same load, a two- wheeled cart does far more damage 
than one with four wheels, and this because of their sudden and 
irregular twisting motion in the trackway. 

Grade Resistance is due to the action of gravity, and is the 
same on good and bad roads. On level roads its effect is immaterial, 
as it acts in a direction perpendicular to the plane of the horizon, and 
neither accelerates nor retards motion. On inclined roads it offers 
considerable resistance, proportional to the steepness of the incline. 



The resistance due to gravity on any incline in pounds per ton 


is equal to ^ r - 

rate ot grade 

The following table shows the resistance due to gravity on dif- 
ferent grades. 

Resistance Due to Gravity on Different Inclinations. 

Grade 1 in 20 30 40 50 60 70 80 90 100 200 300 400 

Rise in feet per mile . . .264 176 132 105 88 75 66 58 52 26 17 13 
Resistance in Ib. per ton . 1 12 74 56 45 38 32 28 25 22 11 J 1\ 5 
The additional resistance caused by inclines may be investigated 
in the following manner: Suppose the whole weight to be borne on 
one pair of wheels, and that the tractive force is applied in a direction 
parallel to the surface of the road. 

Let A B in Fig. 3 represent a portion of the inclined road, C 
being a vehicle just sustained in its position by a force acting in the 
direction C D. It is evident that the vehicle is kept in its position 
by three forces; namely, by its own weight W acting in the vertical 
direction C F, by the force F applied in the direction C D parallel 
to the surface of the road, and by the pressure P which the vehicle 
exerts against the surface of the road acting in the direction C E 

perpendicular to same. To 
determine the relative magni- 
tude of these three forces, 
draw a horizontal line A G 
and the vertical one B G; 
then, since the two lines C F 
and B G are parallel and 
are both cut by the line A B, 

Fig. 3. they must make the two 

angles C F E and A B G 

equal; also the two angles C E F and A G B are equal; therefore, the 
remaining angles F C E and B A G are equal, and the two triangles 
C F E and A B G are similar. And as the three sides of the former 
are proportional to the three forces by which the vehicle is sustained, 
so also are the three sides of the latter; namely, AB or the length 
of the road is proportional to W, or the weight of the vehicle; B G, 



or the vertical rise in the same, to F, or the force required to sustain 
the vehicle on the incline; and A G, or the horizontal distance in 
which the rise occurs, to P, or the force with which the vehicle presses 
upon the surface of the road. Therefore, 

W:AB ::F:GB, 

W :AB : :P : A G. 

If to A G such a value be assigned that the vertical rise of the 
road is exactly one foot, then 


F = - = - = W -sin A 

AB i/AG'+l 

p = W-AG = W-AG 

AB l/AG 2 +l 

in which A is the angle BAG. 

To find the force requisite to sustain a vehicle upon an inclined 
road (the effects of friction being neglected), divide the weight of the 
vehicle and its load by the inclined length of the road, the vertical 
rise of which is one foot, and the quotient is the force required. 

To find the pressure of a vehicle against the surface of an inclined 
road, multiply the weight of the loaded vehicle by the horizontal 
length of the road) and divide the product by the inclined length of 
the same; the quotient is the pressure required. 

The force with which a vehicle presses upon an inclined road 
is always less than its actual weight; the difference is so small that, 
unless the inclination is very steep, it may be taken equal to the 
weight of the loaded vehicle. 

To find the resistance to traction in passing up or down an 
incline, ascertain the resistance on a -level road having the same surface 
as the incline, to which add, if the vehicle ascends, or subtract, if 
it descends, the force requisite to sustain it on the incline; the sum 
or difference, as the case may be, will express the resistance. 

Tractive Power and Gradients. The necessity for easy 
grades is dependent upon the power of the horse to overcome the 
resistance to motion composed of the four forces, friction, collision, 
gravity, and the resistance of the air. 

All estimates on the tractive powder of horses must to a certain 



extent be vague, owing to the different strengths and speeds of animals 
of the same kind, as well as to the extent of their training to any 
particular kind of work. 

The draught or pull which a good average horse, weighing 1,200 
pounds, can exert on a level, smooth road at a speed of 2 miles per 
hour is 100 pounds, equivalent to 22,000 foot-pounds per minute, 
or 13,200,000 foot-pounds per day of 10 hours. 

The tractive power diminishes as the speed increases and, per- 
haps, within certain limits, say from f to 4 miles per hour, nearly 
in inverse proportion to it. Thus the average tractive force of a 
horse, on a level, and actually pulling for 10 hours, may be assumed 

approximately as follows: 

Tractive Power of Horses at Different Velocities. 

Miles per hour. 

Force. Ll>. 

Miles per hour. 

Force. Lb. 


333 33 


Ill 11 



2 i 




90 91 


166 66 


83 33 


142 86 


71 43 




62 50 

The work done by a horse is greatest when the velocity with 
which he moves is f of the greatest velocity with which he can move 
when unloaded; and the force thus exerted is 0.45 of the utmost 
force that he can exert at a dead pull. 

The- traction power of a horse may be increased in about the 
same proportion as the time is diminished, so that when working 
from 5 to 10 hours, on a level, it will be about as shown in the following 



Hours per day Traction (pounds) Hours per day Traction (pounds) 

10 100 7 146| 

9 llli 6 166^ 

8 125 5 200 

The tractive power of teams is about as follows 

1 horse =1 

2 horses 0.95 X 2 - 1.90 

3 " 0.85 X 3 = 2.55 

4 " ' ... 0.80 X 4 - 3.20 




Loss of Tractive Power on Inclines. In ascending in- 
clines a horse's power diminishes rapidly; a large portion of his 
strength is expended in overcoming the resistance of gravity due to 
his own weight and that of the load. Table 5 shows that as the 
steepness of the grade increases the efficiency of both the horse and 
the road surface diminishes; that the more of the horse's energy is 
expended in overcoming gravity the less remains to overcome the 
surface resistance. 


Effects of Grades Upon the Load a Horse can Draw on Different 



Broken Stone. 

Stone Blocks. 







1 :100 





2 :100 





3 100 





4 100 





5 100 





10 100 





15 100 




20 100 



Table 6 shows the gross load which an average horse, weighing 
1,200 pounds, can draw on different kinds of road surfaces, on a 
level and on grades rising five and ten feet per one hundred feet. 


Description of Surface. 


5 per cent 

10 per cent 


13 216 

Broken stone (best condition) 
(slightly muddy) 
' " (ruts and mud) 




" (very bad condition) . . 
Earth (best condition) 
(average condition) 
(moist but not muddy). 
Stone-block pavement (dry and clean) 
" (muddy) 
Sand (wet) 

1 500 



" (dry) 

1 087 



The decrease in the load which a horse can draw upon an incline 
is not due alone to gravity; it varies with the amount of foothold 



afforded by the road surface. The tangent of the angle of inclination 
should not be greater than the coefficient of tractional resistance; 
therefore it is evident that the smoother the road surface, the easier 
should be the grade. The smoother the surface the less the foothold, 
and consequently the load. 

The loss of tractive power on inclines is greater than any inves- 
tigation will show; for, besides the increase of draught caused by 
gravity, the power of the horse is much diminished by fatigue upon 
a long ascent, and even in greater ratio than man, owing to its anatom- 
ical formation and great weight. Though a horse on a level is as 
strong as five men, on a grade of 15 per cent, it is less strong than 
three; for three men carrying each 100 pounds will ascend such a 
grade faster and with less fatigue than a horse with 300 pounds. 

A horse can exert for a short time twice the average tractive 
pull which he can exert continuously throughout the day's work; 
hence, so long as the resistance on the incline is not more than double 
the resistance on the level, the horse will be able to take up the full 
load which he is capable of drawing. 

Steep grades are thus seen to be objectionable, and particularly 
so when a single one occurs on an otherwise comparatively level road, 
in which case the load carried over the less inclined portions must 
be reduced to what can be hauled up the steeper portion. 

The bad effects of steep grades are especially felt in winter, 
when ice covers the roads, for the slippery condition of the surface 
causes danger in descending, as well as increased labor in ascending; 
the water of rains also runs down the road and gulleys it out, destroy- 
ing its surface, thus causing a constant expense for repairs. The 
inclined portions are subject to greater wear from horses ascending, 
thus requiring thicker covering than the more level portions, and 
hence increasing the cost of construction. 

It will rarely be possible, except in a flat or comparatively level 
country, to combine* easy grades with the best and most direct route. 
These two requirements will often conflict. In such a case, increase 
the length. The proportion of this increase will depend upon the 
friction of the covering adopted. But no general rule can be given 
to meet all cases as respects the length which may thus be added, 
for the comparative time occupied in making the journey forms an 




important element in any case which arises for settlement. Disre- 
garding time, the horizontal length of a road may be increased to 
avoid a 5 per cent grade, seventy times the height. 

Table 7 shows, for most practical purposes, the force required 
to draw loaded vehicles over inclined roads. The first column ex- 
presses the rate of inclination; the second, the pressure on the plane 
in pounds per ton; the third, the tendency down the plane (or force 
required to overcome the effect of gravity) in pounds per ton; the 
fourth, the force required to haul one ton up the incline; the fifth, the 
length of level road which would be equivalent to a mile in length of 
the inclined road that is, the length which would require the same 
motive power to be expended in drawing the load over it, as would 
be necessary to draw over a mile of the inclined roadj the sixth, the 
maximum load which an average horse weighing 1,200 pounds can 
draw over such inclines, the friction of the surface being taken at 
-fa of the load drawn. 


Rate of grade. 
Feet per 100 

Pressure on 
the plane in 
Ib. per ton. 

down the 
plane in Ib. 
per ton. 

Power in Ib. 
required to 
haul one ton 
up the plane. 

! length of level 
road. Miles. 

load in Ib. 
which a horse 
can haul. 




45.00 1.000 




- 5.60 

































































































224 . 20 















* Near enough for practice; actually 3239.888. 

Pressure on the plane = weight X nat cos of angle of plane. 

Axle Friction. The resistance of the hub to turning on the 
axle is the same as that of a journal revolving in its bearing, and has 




nothing to do with the condition of the road surface. The coefficient 
of journal friction varies with the material of the journal and its 
bearing, and with the lubricant. It is nearly independent of the 
velocity, and seems to vary about inversely as the square root of the 
pressure. For light carriages when loaded, the coefficient of friction 
is about 0.020 of the weight on the axle; for the ordinary thimble- 
skein wagon when loaded, it is about 0.012. These coefficients are 
for good lubrication; if the lubrication is deficient, the axle friction 
is two to six times as much as above. 

The traction power required to overcome the above axle friction 
for carriages of the usual proportions is about 3 to 3 Ib. per ton of 
the weight on the axle; and for truck wagons, which have medium 
sized wheels and axles, is about 3i. to 4 Ib. per ton. 

Resistance of the Air. The resistance arising from the 
force of the wind will vary with the velocity of the wind, with the 
velocity -of the vehicle, with the area of the surface acted upon, and 
also with the angle of incidence of direction of the wind with the 
plane of the surface. 

The following table gives the force per square foot for various 
velocities : 


Velocity of wind in miles 
per hour. 

Force in Ibs. per sq. ft. 



1 .107 

Pleasant Breeze 



Brisk Gale 



High Wind 




Very High Wind 

50 12.300 Storm 

Effect of Springs on Vehicles. Experiments have shown 
that vehicles mounted on springs materially decrease the resistance to 
traction, and diminish the wear of the road, especially at speeds 
beyond a walking pace. Going at a trot, they were found not to 
cause more wear than vehicles without springs at a, walk, all other 
conditions being similar. Vehicles with springs improperly fixed 
cause considerable concussion, which in turn destroys the road 




The considerations governing the location of country roads are 
dependent upon the commercial condition of the country to be 
traversed. In old and long-inhabited sections the controlling ele- 
ments will be the character of the traffic to be accommodated. In 
such a section, the route is generally predetermined, and therefore 
there is less liberty of a choice and selection than in a new and sparsely 
settled district, where the object is to establish the easiest, shortest, 
and most economical line of intercommunication according to the 
physical character of the ground. 

Whichever of these two cases may have to be dealt with, the same 
principle governs the engineer, namely, to so lay out the road as to 
effect the conveyance of the traffic with the least expenditure of 
motive power consistent with economy of construction and main- 

Economy of motive power is promoted by easy grades, by the 
avoidance of all unnecessary ascents and descents, and by a direct 
line; but directness must be sacrificed to secure easy grades and to 
avoid expensive construction. 

Reconnoissance. The selection of the best route demands 
much care and consideration on the part of the engineer. To obtam 
the requisite data upon which to form his judgment, he must make 
a personal reconnoissance of the district. This requires that the 
proposed route be either ridden or walked over and a careful examina- 
tion made of the principal physical contours and natural features of 
the district. The amount of care demanded and the difficulties 
attending the operations will altogether depend upon the character 
of the country. 

The immediate object of the reconnoissance is to select one or 
more trial lines, from which the final route may be ultimately deter- 

When there are no maps of the section traversed, or when those 
which can be procured are indefinite or inaccurate, the work of 
reconnoitering will be much increased. 

In making a reconnoissance there are several points which, if 
carefully attended to, will very considerably lessen the labor and 
time otherwise required. Lines which would run along the imme- 



diate bank of a large stream must of necessity intersect all the tribu- 
taries confluent on that bank, thereby demanding a corresponding 
number of bridges. Those, again, which are situated along the 
slopes of hills are more liable in rainy weather to suffer from washing 
away of the earthwork and sliding of the embankments; the others 
which are placed in valleys or elevated plateaux, when the line crosses 
the ridges dividing the principal water courses will have steep ascents 
and descents. 

In making an examination of a tract of country, the first point 
to attract notice is the unevenness or undulations of its surface, which 
appears to be entirely without system, order, or arrangement; but 
upon closer examination it will be perceived that one general prin- 
ciple of configuration obtains even in the most irregular countries. 
The country is intersected in various directions by main water courses 
or rivers, which increase in size as they approach the point of their 
discharge. Towards these main rivers lesser rivers approach on 
both sides, running right and left through the country, and into these, 
again, enter still smaller streams and brooks. The streams thus 
divide the hills into branches or spurs having approximately the same 
direction as themselves, and the ground falls in every direction from 
the main chain of hills towards the water courses, forming ridges 
more or less elevated. 

The main ridge is cut down at the heads of the streams into 
depressions called gaps or passes; the more elevated points are called 
peaks. The water which has fallen upon these peaks is the origin 
of the streams which have hollowed out the valleys. Furthermore, 
the ground falls in every direction towards the natural water courses, 
forming ridges more or less elevated running between them and 
separating from each other the districts drained by the streams. 

The natural water courses mark not only the lowest lines, but 
the lines of the greatest longitudinal slope in the valleys through which 
they flow. 

The direction and position of the principal streams give also 
the direction and approximate position of the high ground or ridges 
which lie between them. 

The positions of the tributaries to the larger stream generally 
indicate the points of greatest depression in the summits of the ridges, 



Fig. 4. Contour Lines. 



and therefore the points at which lateral communication across the 
high ground separating contiguous valleys can be most readily made. 

The instruments employed in reconnoitering, are : The compass, 
for ascertaining the direction; the aneroid barometer, to fix the ap- 
proximate elevation of summits, etc. ; and the hand level, to ascertain 
the elevation of neighboring points. If a vehicle can be used, an 
odometer may be added, but distances can usually be guessed or 
ascertained by time estimates or otherwise, closely enough for pre- 
liminary purposes. The best maps obtainable and traveling com- 
panions who possess a local knowledge of the country, together with 
the above outfit is all that will be necessary for the first inspection. 

The reconnoissance being completed, instrumental surveys of 
the routes deemed most advantageous should be made. When the 
several lines are plotted to the same scale, a good map can be pre- 
pared from which the exact location of the road can be determined. 

In making the preliminary surveys the topographical features 
should be noted for a convenient distance to the right and the left of 
the line, and all prominent points located by compass bearings. The 
following data should also be obtained: the importance, magnitude, 
and direction of all streams and roads crossed; the character of the 
material to be excavated or available for embankments, the position 
of quarries and gravel pits, and the modes of access thereto; and all 
other information that may effect a selection. 

Topography. There are various methods of delineating upon 
paper the irregularities of the surface of the ground. The method 
of most utility to the engineer is that by means of "contour lines." 
These are fine lines traced through the points of equal level over the 
surface surveyed, and denote that the level of the ground throughout 
the whole of their course is identical; that is to say, that every part 
of the ground over which the line passes is at a certain height above 
a known fixed point termed the datum, this height being indicated 
by the figures written against the line. 

The intervals between the lines vertically are equal and. may 
be.l, 3, 5, 10 or more feet apart; where the surface is very steep they 
lie close together. These lines by their greater or less distance apart 
have the effect of shading, and make apparent to the eye, the 
undulations and irregularities in the surface of the country. 





- 54.4- 

- 54.62 

Fig. 4 shows an imaginary tract of country, the physical features 
of which are shown by contour lines. 

Map. The map should show the 
lengths and direction of the different por- 
tions of the line, the topography, rivers, 
water courses, roads, railroads, and other 
matters of interest, such as town and 
county lines, dividing lines between property, 
timbered and cultivated lands, etc. 

Any convenient scale may be adopted; 
400 feet to an inch will be found the most 

Memoir. The descriptive memoir 
should give with minuteness all information, 
such as the nature of the soil, character of 
the several excavations whether earth or 
rock, and such particular features as can- 
not be clearly shown upon the map or 

Special information should be given re- 
garding the rivers crossed, as to their width, 
depth at highest known flood, velocity of 
current, character of banks and bottom, 
and the angle of skew which the course 
makes with the line of the road. 

Levels. Levels should be taken along 
the course of each line, usually at every 100 
feet, or at closer intervals, depending upon 
the nature of the country. 

In taking the levels, the heights of 
all existing roads, railroads, rivers, or 
canals should be noted. "Bench marks" 
should be established at least every half 
mile, that is, marks made on any fixed 
object, such as a gate post, side of a house, 
or, in the absence of these, a cut made 
on a large tree. The height and exact 




Fig. 5. Preliminary Profile 



description of each bench mark should be recorded in the level book. 

Cross Levels. Wherever considered necessary levels at right 
angle to the center line should be taken. These will be found useful 
in showing what effect a deviation to the right or left of the surveyed 
line would have. Cross levels should be taken at the intersection of 
all roads and railroads to show to what extent, if any, these levels 
will have to be altered to suit the levels of the proposed road. 

Profile. A profile is a longitudinal section of the route, made 
from the levels. Its horizontal scale should be the same as that of 
the map; the vertical scale should be such as will show with distinct- 
ness the inequalities of .the ground. ' 

Fig. 5 shows the manner in which a profile is drawn and the 
nature of the information to be given upon it. 

Bridge Sites. The question of choosing the site of bridges is 
an important one. If the selection is not restricted to a particular 
point, the river should be examined for a considerable distance above 
and below what would be the most convenient point for crossing; and 
if a better site is found, the line of the road must be made subordinate 
to it. If several practicable crossings exist, they carefully 
compared in order to select the one most advantageous. The follow- 
ing are controlling conditions: (1) Good character of the river bed, 
affording a firm foundation. If rock is present near the surface of 
the river bed, the foundation will be easy of execution and stability 
and economy will be insured. (2) Stability of river banks, thus 
securing a permanent concentration of the waters in the same bed. 
(3) The axis of the bridge should be at right angles to the direction 
of the current. (4) Bends in rivers are not suitable localities and 
should be avoided if possible. A straight reach above the bridge 
should be 'secured if possible. 

Final Selection. In making the final selection the following 
principles should be observed as far as practicable. 

(a) To follow that route which affords the easiest grades. The 
easiest grade for a given road will depend on the kind of covering 
adopted for its surface. 

(6) To connect the places by the shortest and most direct route 
commensurate with easy grades. 

(c) To avoid all unnecessary ascents and descents. When a 



road is encumbered with useless ascents, the wasteful expenditure of 
power is considerable. 

(d) To give the center line such a position, with reference to 
the natural surface of the ground, that the cost of construction shall 
be reduced to the smallest possible amount. 

(e) To cross all obstacles (w T here- structures are necessary) as 
nearly as possible at right angles. The cost of skew structures 
increases nearly as the square of the secant of the obliquity. 

(/) To cross ridges through the lowest pass which occurs. 

(gr) To cross either under or over railroads; for grade crossings 
mean danger to every user of the highway. 

Examples of Cases to be Treated. In laying out the line 
of a road, there are three cases which may have to be treated, and 
each of these is exemplified in the contour map, Fig. 4. First, the 
two places to be connected, as the towns A and B on the plan, may 
be both situated in the same valley, and upon the same side of it ; that 
is, they are not separated from each other by the main stream which 
drains the valley. This is the simplest case. Secondly, although 
both in the same valley, the two places may be on opposite sides of 
the valley, as at A and C, being separated by the main river. Thirdly, 
they may be situated in different valleys, separated by an intervening 
ridge of ground more or less elevated, as at A and D. In laying out 
an extensive line of road, it frequently happens -that all these cases 
have to be dealt with. 

The most perfect road is that of which the course is perfectly 
straight and the surface practically level; and, all other things being 
the same, the best road is that which answers nearest to this de- 

Now, in the first case, that of the two towns situated on the 
same side of the main valley, there are two methods which may be 
pursued in forming a communication between them. A road follow- 
ing the direct line between them, shown by the thick dotted line A B, 
may be made, or a line may be adopted which will gradually and 
equally incline from one town to another, supposing them to be at 
different levels; or, if they are on the same level, the line should keep 
at that level throughout its entire course, following all the sinuosities 
and curves which the irregular formation of the country may render 



necessary for the fulfillment of these conditions. According to the 
first method, a level or uniformly inclined road might be made from 
one to the other; this line would cross all the valleys and streams 
which run down to the main river, thus necessitating deep cuttings, 
heavy embankments, and numerous bridges; or these expensive 
works might be avoided by following the sinuosities of the valley. 
When the sides of the main valley are pierced by numerous ravines 
with projecting spurs and ridges intervening, instead of following the 
sinuosities, it will be found better to make a nearly straight line 
cutting through the projecting points in such a way that the material 
excavated should be just sufficient to fill the hollows. 

Of all these, the best is the straight or uniformly inclined, or 
level road, although at the same time it is the most expensive. If 
the importance of the traffic passing between the places is not suffi- 
cient to warrant so great an outlay, it will become a matter of consider- 
ation whether the course of the road should be kept straight, its surface 
being made to undulate with the natural face of the country; or 
whether,' a level or equally inclined line being adopted, the course 
of the road should be made to deviate from the direct line, and follow 
the winding course which such a condition is supposed to necessitate. 

In the second case, that of two places situated on opposite sides 
of the same valley, there is, in like manner, the choice of a perfectly 
straight Hne to connect them, which would probably require a big 
embankment if the road was kept level, or steep inclines if it followed 
the surface of the country; or by winding the road, it may be carried 
across the valley at a higher point, where, if the level road be taken, 
the embankment would not be so high, or, if kept on the surface, 
the inclination would be reduced. 

In the third case, there is, in like manner, the alternative of 
carrying the road across the intervening ridge in a perfectly straight 
line, or of deviating it to the right and left, and crossing the ridge 
at a point where the elevation is less. 

The proper determination of the question which of these courses 
is the best under certain circumstances involves a consideration of 
the comparative advantages and disadvantages of inclines and 
curves. What additional increase in the length of a road would be 
equivalent to a given inclined plane upon it; or conversely, what 


inclination might be given to a road as an equivalent to a given de- 
crease in its length ? To satisfy this question, the comparative force 
required to draw different vehicles with given loads must be known, 
both upon level and variously inclined roads. 

The route which will give the most general satisfaction consists 
in following the valleys as much as possible and rising afterward by 
gentle grades. This course traverses the cultivated lands, regions 
studded with farmhouses and factories. The value of such a line 
is much more considerable than that of a route by the ridges. The 
water courses which flow down to the main valley are, it is true, 
crossed where they are the largest and require works of large dimen- 
sions, but also they are fewer in number. 

Intermediate Towns. Suppose that it is desired to form a 
road between two distant towns, A and B, Fig. 6, and let us for the 
present neglect altogether the consideration of the physical features 
of the intervening country, assuming that it is equally favorable 
whichever line we select. Now at first sight, it would appear that 
under such circumstances a perfectly straight line drawn from one 

town to the other would be 

"X the best that could be chos- 

N N en. On more careful exam- 

N N ination however, of the lo- 

S N N cality, we may find that 

x s v there is a third town, C, 

situated somewhat on one 
side of the straight line 

which we have drawn from A to B ; and although our primary object 
is to connect only the two latter, that it would nevertheless be of 
considerable service if the whole of the three towns were put into 
mutual connection with each other. 

This may be effected in three different ways, any one of which 
might, under the circumstances, be the best. In the first place, we 
might, as originally suggested, form a straight road from A to B, 
and in a similar manner two other straight roads from A to C, and 
from B to C, and this would be the most perfect way of effecting the 
object in view the distance between any of the two towns being 
reduced to the least possible. It would, however, be attended with 



considerable expense, and it would he requisite to construct a much 
greater length of road than according to the second plan, which would 
be to form, as before, a straight road from A to B, and from C to con- 
struct a road which should join the former at a point D, so as to be per- 
pendicular to it. The traffic between A or B and C would proceed to 
the point D and then turn off to C. With this arrangement, white 
the length of the roads would be very materially decreased, only a 
slight increase would be occasioned in the distance between C and 
the other two towns. The third method would be to form only the 
two roads A C and C B, in which case the distance between A and B 
would be somewhat increased, while that between A C or B and C 
would be diminished, and the total length of road to be constructed 
would also be lessened. 

As a general rule it may be taken that the last of these methods 
is the best and most convenient for the public; that is to say, that 
if the physical character of the country does not determine the course 
of the road, it will generally be found best not to adopt a perfectly 
straight line, but to vary the line so as to pass through all the prin- 
cipal towns near its general course. 

flountain Roads. The location of roads in mountainous 
countries presents greater difficulties than in an ordinary undulating 
country; the same latitude in adopting undulating grades and choice 
of position is not permissible, for the maximum must be kept before 
the eye perpetually. A mountain road has to be constructed on the 
maximum grade or at grades closely approximating it, and but one 
fixed point can be obtained before commencing the survey, and that 
is the lowest pass in the mountain range; from this point the survey 
must be commenced. The reason for this is that the lower slopes 
of the mountain are flatter than those at their summit; they cover a 
larger area, and merge into the valley in diverse undulations. So 
that a road at a foot of a mountain may be carried at will in the 
desired direction by more than one route, while at the top of a moun- 
tain range any deviation from the lowest pass involves increased 
length of line. The engineer having less command of the ground, 
owing to the reduced area he has to deal with and the greater abrupt- 
ness of the slopes, is liable to be frustrated in his attempt to get his 
line carried in the desired direction. ' 



It is a common practice to run a mountain survey up hill, but 
this should be avoided. Whenever an acute-angled zigzag is met 
with on a mountain road near the summit, the inference to be drawn 
is that the line being carried up hill on reaching the summit was 
too low and the zigzag was necessary to reach the desired pass. The 
only remedy in such a case is by a resurvey beginning at the summit 
and running down hill. This method requires a reversal of that 
usually adopted. The grade line is first staked out and its horizontal 
location surveyed afterwards. The most appropriate instrument for 
this work is a transit with a vertical circle on which the telescope may 
be set to the angle of the maximum grade. 

Loss of Height. Loss of height 'is to be carefully avoided in a 
mountain road. By loss of height is meant an intermediate rise in a 
descending grade. If a descending grade is interrupted by the intro- 
duction of an unnecessary ascent, the length of the road will be in- 
creased over that due to the continuous grade by the length of the 
portion of the road intervening between the summit of the rise and 
the point in the road on a level with that rise a length which is double 
that due on the gradient to the height of the rise. For example, 
if a road descending a mountain rises at some intermediate point to 
cross over a ridge or spur, and the height ascended amounts to 110 
feet before the descent is continued, such a road would be just one 
mile longer than if the descent had been uninterrupted; for 110 feet 
is the rise due to a half-mile length at 1 : 24. 

Water on Mountain Roads. Water is needed by the work- 
men and during the construction of the road ; it is also very necessary 
for the traffic, especially during hot weather; and if the road exceeds 
5 miles in length, provision should be made to have it either close 
to or within easy reach of the road. With a little ingenuity the 
water from springs above the road, if such exist, can be led down to 
drinking fountains for men, and to troughs for animals. 

In a tropical country it would be a matter for serious consider- 
ation if the best line for a mountain road 10 miles in length or up- 
wards, but without water, should not be abandoned in favor of a 
worse line with a water supply available. 

Halting Places. On long lines of mountain roads halting 
places should be provided at frequent intervals. 


Alignment. No rule can be laid down for the alignment of a 
road; it will depend both upon the character of the traffic on it and 
upon the "lay of the land." To promote economy of transportation 
it should be straight; but if s-traightness is obtained at the expense 
of easy grades that might have been obtained by deflections and 
increase in length, it will prove very expensive to the community 
that uses it. 

Where curves are necessary, employ the greatest radius possible 
and never less than fifty feet. They may be circular or parabolic. 
The parabolic will be found exceedingly useful for joining tangents 
of unequal length, and for following contour lines; its curvature 
being least at its beginning and* ending, makes the deviations from 
a straight line less strongly marked than by a circular arc. 

When a curve occurs on an ascent, the grade at that place must 
be diminished in order to compensate for the additional resistance of 
the curve. 

The width of the wheel way on curves must be increased. This 
increase should be one-quarter of the width for central angles between 
90 and 120 degrees, and one-half for angles between 60 and 90 degrees. 
Excessive crookedness of alignment is to be avoided, for any unneces- 
sary length causes a constant threefold waste; first, of the interest 
of the capital expended in making that unnecessary portion; secondly, 
of the ever recurring expense of repairing it; and thirdly, of the time 
and labor employed on travelling over it. 

.The curving road around a hill may be often no longer than the 
straight one over it, for the latter is straight only with reference to 
the horizontal plane, while it is curved as to the vertical plane; -the 
former is curved as to the horizontal plane, but straight as to the 
vertical plane. Both lines curve, and we call the one passing over 
the hill straight only because its vertical curvature is less apparent 
to our eyes. 

The differen.ce in length between a straight road and one which 
is slightly curved is very small. If a road between two places ten 
miles apart were made to curve so that the eye could nowhere see 
farther than one-quarter of a mile of it at once, its length would 
exceed that of a straight road between the same points by only about 
four hundred and fifty feet. 



Zigzags. The method of surmounting a height by a series of 
zigzags or by a series of reaches with practicable curves at the turns, 
is objectionable. 

(1) An acute-angled zigzag obliges the traffic to reverse its 
direction without affording it convenient room for the purpose. The 
consequence is that with slow traffic a single train of vehicles is 
brought to a stand, while if two trains of vehicles travelling in opposite 
directions meet at the zigzag a block ensues. 

(2) With zigzags little progress is made towards the ultimate 
destination of the road ; height is surmounted, but horizontal distance 
is increased for which there is no necessity or compensation. 

(3) Zigzags are dangerous. In case of a runaway down hill 
the zigzag must prove fatal. 

(4) If the drainage cannot be carried clear of the road at the 
end of each reach, it must be carried under the road in one reach only 
to appear again at the next, when a second bridge, culvert, or drain 
will be required, and so on at the other reaches. If the drainage can 
be carried clear at the termination of each reach, the lengths between 
the curves will be very short, entailing numerous zigzag curves, which 
are expensive to construct and maintain. 

Final Location. The route being finally determined upon, it 
requires to be located. This consists in tracing the line, placing a 
stake at every 100 feet on the straight portions and at every 50 or 

25 feet on the curves. At the tangent point of curves, and at points 
of compound and reverse curves, a larger and more permanent stake 
should be placed. Lest those stakes should be disturbed in the 
process of construction, their exact distance from several points 
outside of the ground to be occupied by the road should be carefully 
measured and recorded in the notebook, so that they may be replaced. 
The stakes above referred to show the position of the center line of 
the road, and form the base line from which all operations of con- 
struction are carried on. Levels are taken at each stake," and cross 
levels are taken at every change of longitudinal slope. 

Construction Profile. The construction or working profile 
is made from the levels obtained on location. It should be drawn to a 
horizontal scale of 400 feet to the inch and a vertical scale of 20 feet 
to the inch. Fig. 7 represents a portion of such a profile. The 




figures in column A represent the elevation of the ground at every 
100 feet, or where a stake has been driven, above datum. The 
figures in column B are the elevations of the grade above datum. 
The figures in column C indicate the depth of cutting or height of 
fill; they are obtained by taking the difference between the level of 
the road and the level of the surface of the ground. The straight line 



2.50.' 100 

P . o o ' in x 7 7 I ^00:|00 * * i \/ 
OoQSoJjo'o'oo l 

M ? ? i : i i ? i ? ? 1 5 1 i 


A g i ) a 


SSS lr '' v '>in]'nS^<I>^?i 
<oa>coooi--og N fflKioin!o 



at the top represents the grade of the road; the upper surface of the 
road when finished would be somewhat higher than this, while the 
given line represents what is termed the sub-grade or formation level. 
All the dimensions refer to the formation level, to which the surface 
of the ground is to be formed to receive the road covering. 

At all changes in the rate of inclination of the grade line a heavier 
vertical line should be drawn. 

Gradient. The grade of a line is its longitudinal slope, and 
is designated by the proportion between its length and the difference 
of height of its two extremes. The ratio of these two qualities gives 
it its name; if the road ascends or falls one foot in every twenty feet 
of its length, it is said to have a grade of 1 : 20 or a 5 per cent grade. 
Grades are of two kinds, maximum and minimum. The maximum 
is the steepest which is to be permitted and which on no account is to be 
exceeded. The minimum is the least allowable for good drainage. 
(For method of designating grades see Table 9). 

Determination of Gradients. The maximum grade is fixed 
by two considerations, one relating to the power expended in ascend- 
ing, the other to the acceleration in descending the incline. 

There is a certain inclination, depending upon the degree of 
perfection given to the surface of the road, which cannot be exceeded 



without a direct loss of tractive power. This inclination is that in 
descending which, at a uniform speed, the traces slacken, or which 
causes the vehicles to press on the horses; the limiting inclination 
within which this effect does not take place is the angle of repose. 


American method. 
Feet per 100 feet. 

English method. 

Feet per mile. 

Angle with the horizon. 

\ 1 400 






































2i ] 











36 145.2 




33| 158.4 




30f 171.6 




28$ 184.8 




26| 198 





25 211.2 


1 7 



23* 221.4 

















































1 10 





The angle of repose for any given road surface can be easily 
ascertained from the tractive force required upon a level with the 
same character of surface. Thus if the force necessary on a level 
to overcome the resistance of the load is ^ f l ^ weight, then the 
same fraction expresses the angle of repose for that surface. 

On all inclines less steep than the angle of repose a certain 
amount of tractive force is necessary in the descent as well as in 
the ascent, and the mean of the two drawing forces, ascending and 
descending, is equal to the force along the level of the road. Thus 
on such inclines, as much mechanical force is gained in the descent 
as is lost in the ascent. From this it inferred that when a 
vehicle passes alternately each way along the road, no real loss is 



occasioned by the inclination of the road; such is not, however, 
practically the fact with animal power, for while it is necessary in 
the ascending journey to have either a less or a greater number of 
horses than would be requisite if the road were entirely level, no 
corresponding reduction can be made in the descending journey. 
On inclines which are more steep than the angle of repose, the load 
presses on the horses during their descent, so as to impede their 
action, and their power is expended in checking the descent of the 
load; or if this effect be prevented by the use of any form of drag or 
brake, then the power expended on such a drag or brake corresponds 
to an equal quantity of mechanical power expended in the ascent, 
for which no equivalent is obtained in the descent. 

The maximum grade for a given road will depend (1) upon the 
class of traffic that will use it, whether fast and light, slow and heavy, 
or mixed, consisting of both light and heavy; (2) upon the character 
of the pavement adopted; and (3) upon the question of cost of con-' 
struction. Economy of motive power and low cost of construction are 
antagonistic to each other, and the engineer will have to weigh the 
two in the balance. 

For fast and light traffic the grades should not exceed 2 per 
cent; for mixed traffic 3 per cent may be adopted; while for slow 
traffic combined with economy 5 per cent should not be exceeded. 
This grade is practicable but not convenient. 

Minimum Grade. From the previous considerations it would 
appear that an absolutely level road was the one to be sought for, but 
this is not so; there is a minimum or least allowable grade which the 
road must not fall short of, as well as a maximum one which it must 
not exceed. If the road was perfectly level in its longitudinal direc- 
tion, its surface could not be kept free from water without giving it 
so great a rise in its middle as would expose vehicles to the danger of 
overturning. The minimum grade commonly used is 1 per cent. 

Undulating Grades. From the fact that the power required 
to move a load at a given velocity on a level road is decreased on a 
descending grade to the same extent it is increased in ascending the 
same grade, it must not be inferred that the animal force expended 
in passing alternately each way over a rising and falling road will 
gain as much in descending the several inclines as it will lose in ascend- 



ing them. Such is not the case. Theanimal force must be sufficient, 
either in power or number, to draw the load over the level portions 
and up the steepest inclines of the road, and in practice no reduction 
in the number of horses can be made to correspond with the decreased 
power required in descending the inclines. 

The popular theory that a gentle undulating road is less fatiguing 
to horses than one which is perfectly level is erroneous. The asser- 
tion that the alternations of ascent, descent, arid levels call into play 
different muscles, allowing some to rest while others are exerted, 
and thus relieving each in turn, is demonstrably false, and con- 
tradicted by the anatomical structure of the horse. Since this doc- 
trine is a mere popular error, it should be utterly rejected, not only 
because false in itself, but still more because it encourages the building 
of undulating roads, and this increases the labor and cost of trans- 
portation upon them. 

Level Stretches. On long ascents' it is generally recom- 
mended to introduce level or nearly level stretches at frequent inter- 
vals in order to rest the animals. These are objectionable when 
they cause loss of height, and animals will be more rested by halting 
and unharnessing for half an hour than by travelling over a level 
portion. The only case which justifies the introduction of levels 
into an ascending road is where such levels will advance the road 
towards its objective point; where this is the case there will be no 
loss of either length or height, and it will simply be exchanging a 
level road below for a level road above. 

Establishing the Grade. . When the profile of a proposed 
route has been made, a grade line is drawn upon it (usually in red) in 
such a manner as to follow its general slope, but to average its irregular 
elevation and depressions. 

If the ratio between the whole distance and the height of the line 
is less than the maximum grade intended to be used, this line will be 
satisfactory; but if it be found steeper, the cuttings or the length 
of the line will have to be increased; the latter is generally preferable. 

The apex or meeting point of all curves should be rounded off 
by a vertical curve, as shown in Fig. 8, thus slightly changing the 
grade at and near the point of intersection. A vertical curve rarely 
need extend more than 200 feet each way from that point. 



Let A B, B C, be two grades in profile, intersecting at station B, 
and let A and C be the adjacent stations. It is required to join the 

Fig. 8. 

grades by a vertical curve extending from A to C. Imagine a chord 
drawn from A to C. The elevation of the middle point of the chord 
will be a mean of the elevations of the grade at A and C, and one- 
half of the difference" between this and the elevation of the grade at 
B will be the middle ordinate of the curve. Hence we have 

M = - 

in which M equals the correction in grade for the point B. The 
correction for any other point is proportional to the square of its 
distance from A or C. Thus the correction A+ 25. is T VM; at 
A + 50 it is \ M; at A + 75 it is T 9 6 M; and the same for corre- 
sponding points on the other side of B. The corrections in this case 
shown are subtractive, since M is negative. They are additive 
when M is positive, and the curve concave upward. 


A road should be wide enough to accommodate the traffic for 
which it is intended, and should comprise a wheelway for vehicles 
and a space on each side for pedestrians. 

The wheelway of country highways need be no wider than is 
absolutely necessary to accommodate the traffic using it; in many 
places a track wide enough for a single team is all that is necessary. 
But the breadth of the land appropriated for highway purposes 
should be sufficient to provide for all future increase of traffic. The 
wheelways of roads in rural sections should be double; that is, one 
portion paved (preferably the center), and the other left with the 



natural soil. The latter if kept in repair will for at least one-half 
the year be preferred by teamsters. 

The minimum width of the paved portion, if intended to carry 
two lines of travel, is fixed by the width required to allow two vehicles 
to pass each other safely. This width is 16 feet. If intended for 
a single line of travel, 8 feet is sufficient, but suitable turnouts must be 
provided at frequent intervals. The most economical width for any 
roadway is some multiple of eight. 

Wide roads are the best; they expose a larger surface to the 
drying action of the sun and wind, and require less supervision than 
narrow ones. Their first cost is greater than narrow ones, and that 
nearly in the ratio of the increased width. 

The cost of maintaining a mile of road depends more upon the 
extent of the traffic. than upon the extent of its surface, and unless 
extremes be taken, the same quantity of material will be necessary 
for the repair of the road whether wide or narrow, which is subjected 
to the same amount of traffic. The cost of spreading the materials 
over the wide road will be somewhat greater, but the cost of the 
materials will be the same. On narrow roads the traffic, being 
confined to one track, will wear more severely than if spread over a 
wider surface. 

The width of land appropriated for road purposes varies in the 
United States from 49^ feet to 66 feet; in England and France from 
26 to 66 feet. And the width or space macadamized is also subject 
to variation; in the United States the average width is 16 feet; in 
France it varies between 16 and 22 feet; in Belgium 8J feet seems 
to be the regular width, while in Austria from 14^ to 26 J feet. 

Transverse Contour. The center of all roadways should 
be higher than the sides. The object of this is to facilitate the flow 
of the rain water to the gutters. Where a good surface is. maintained 
a very moderate amount of rise is sufficient for this purpose. Earth 
roads require the most and asphalt the least. The rise should bear 
a certain proportion to the width of the carriageway. The most 
suitable proportions for the different paving materials is shown in 
table 10. 

Form of Transverse Contour. All authorities agree that 
the form should be convex, but they differ in the amount and form 


of the convexity. Circular arcs, two straight lines joined by a circular 
arc, and ellipses, all have their advocates. 

TABLE 10. 

Kind of Surface. Proportions of the 

Carriageway. Width. 

Earth Rise at center fo 

Gravel -fa 

Broken Stone fa 

For country roads a curve of suitable convexity may be obtained 
as follows: Give of the total rise at \ the width from the center 
to the side, and f of the total rise at ^ the width (Fig. 9). 

Excessive height and convexity of cross-section contract the 
width of the wheelway, by concentrating the traffic at the center, 
that being the only part where a vehicle can run upright. The force' 
required to haul vehicles over such cross-sections is increased, be- 

cause an undue proportion of the load is thrown upon two wheels 
instead of being distributed equally over the four. The continual 
tread of horses' feet in one track soon forms a depression which holds 
water, and the surface is not so dry as with a flat section, which allows 
the traffic to distribute itself over the whole width. 

Sides formed of straight lines are also objectionable. They 
wear hollow, retain water, and defeat the object sought by raising 
the center. 

The required convexity should be obtained by rounding the 
formation surface, and not by diminishing the thickness of the 
covering at the sides. 

Although on hillside and mountain roads it is generally recom- 
mended that the surface should consist of a single slope inclining 
inwards, there is no reason for "or advantage gained by this method. 
The form best adapted to these roads is the same as for a road under 
ordinary conditions. 

With a roadway raised in the center and the rain water draining 
off to gutters on each side, the drainage will be more effectual and 



speedy than if the drainage of the outer half of the road has to pass 
over the inner half. The inner half of such a road is usually sub- 
jected to more traffic than the outer half. If formed of a straight 
incline, this side will be worn hollow and retain water. The inclined 
flat section never can be properly repaired to withstand the traffic. 
Consequently it never can be kept in good order, no matter how 
constantly it may be mended. It is always below par and when 
heavy rain falls it is seriously damaged. 

In the construction of roads, drainage is of the first importance. 
The ability of earth to sustain a load depends in a large measure upon 
the amount of moisture retained by it. Most earths form a good 
firm foundation so long as they are kept dry, but when wet they lose 
their sustaining power, becoming soft and incoherent. 

The drainage of roadways is of two kinds, viz., surface and sub- 
surface. The first provides for the speedy removal of all water 
falling on the surface of the road ; the second provides for the removal 
of the underground water found in the body of the road, a thorough 
removal of which is of the utmost importance and essential to the 
life of the road. A road covering placed on a wet undrained bottom 
will be destroyed by both water and frost, and will always be trouble- 
some and expensive to maintain; perfect subsoil drainage is a neces- 
sity and will be found economical in the end even if in securing it 
considerable expense is required. 

The methods employed for securing the subsoil drainage must 
be varied according to the character of the natural soil, each kind of 
soil requiring different treatment. 

The natural soil may be divided into the following classes: 
silicious, argillaceous, and calcareous; rock, swamps, and morasses. 

The silicious and calcareous soils, the sandy loams and rock, 
present no great difficulty in securing a dry and solid foundation. 
Ordinarily they are not retentive of water and therefore require no 
underdrains; ditches on each side of the road will generally be found 

The argillaceous soils and softer marls require more care; they 
retain water and are difficult to compact, except at the surface; 
and they are very unstable under the action of water and frost. 



The drainage of these soils may be effected by transverse drains 
and deep side ditches of ample width. The transverse drains are 
placed across the road, not at right angles but in the form of an 
inverted V with the point directed up hill; the depth at the angle 
point should not be less than 18 inches below the subgrade surface, 
and each branch should descend from the apex to the side ditches 
with a fall of not less than 1 inch in 5 feet. The distance apart of 
these drains will depend upon the wetness of the soil; in the case of 
very w r et soil they should be at intervals of 15 feet, which may be 
increased to 25 feet as the ground becomes drier and firmer. 

The transverse drains are best formed of unglazed circular tile 
of a diameter not less than 3 inches, jointed with loose collars. The 
tiles are made from terra cotta or burnt clay, are porous, and are 
superior to all other kinds of drains. They carry off the water with 
greater ease, rarely if ever get choked up, and only require a slight 
inclination to keep the water moving through them. 

Fig. 10. Tile Drain. 

Fig. 11. Silt Basin. 

The tiles are made in a variety of forms, as horseshoe, sole, 
double sole, and round, the name being derived from the shape of 
the cross-sections. Round tile is superior to all other forms. The 
inside diameter of these tiles varies from 1 to 6 inches, but they are 
manufactured as large as 24 inches. Pieces of the larger pipe serve 
as collars for the smaller ones. They are made in lengths of 12, 
14 and 24 inches, and in thickness of shell from j'of an inch to 1 inch. 

The collar which encircles the joint of the small tile allows a 
large opening, and at the same time prevents sand and silt from 



entering the drain. Perishable material should not be used for 
jointing. When laid in the ditch they should be held in place by 
small stones. Connections should be made by proper Y-branches. 
The outlets may be formed by building a dwarf wall of brick or 
stone, whichever is the cheapest or most convenient in the locality. 
The outlet should be covered with an iron grating to prevent vermin 
entering the drain pipes, building nests and thus choking up the 
waterway. (See Fig. 12.) 

Fig. 12. Outlet. 

Silt-basins should be constructed at all junctions and wherever 
else they may be considered necessary; they may be made from a 
single 6-inch pipe (Fig. 11) or constructed of brick masonry. 

The trenches for the tile should be excavated at least 3 feet 
wide on top and 12 inches on the bottom. After the tiles are laid 
the trenches must be filled to subgrade level with round field or 
cobble stones; stones with angular edges are unsuitable for this 
purpose. Fine gravel, sand, or soil should not be placed over the 
drains. Bricks and flat stones may be substituted for the tiles, 
and the trenches filled as above stated. 

As tile drains are more liable to injury from frost than those 
of either brick or stone, their ends at the side ditches should not 
in very cold climates be exposed directly to the weather, but may 
terminate in blind drains, or a few lengths of vitrified clay pipe 
reaching under the road a distance of about 3 to 4 feet from the 
inner slope of the ditch. 

Another method of draining the roadbed offering security from 
frost is by one or more rows of longitudinal drains. These drains 
are placed at equal distances from the side ditches and from each 
other, and discharge into cross drains placed from 2oO to 300 feet 


apart, more or less, depending on the contour of the ground. The 
cross drains into which they discharge should be of ample dimensions. 
On these longitudinal lines of tiles the introduction of catch basins 
at intervals of 50 feet will facilitate the removal of the water. These 
catch basins may be excavated three or more feet square and as deep 
as the tiles are laid. After the tiles are laid the pit is filled with gravel 
and small stones. 

Fig. 13. 

Fall of Drains. It is a mistake to give too much fall to small 
drains, the only effect of which is to produce such a current through 
them as will wash away or undermine the ground around them, and 
ultimately cause their own destruction. When a drain is once closed 
by any obstruction no amount of fall which could be given it will 

Fig. 14. 

again clear the passage. A drain with a considerable current through 
it is much more likely to be stopped from foreign matter carried into 
it, which a less rapid stream could not have transported. 

A fall of 1 inch in 5 feet will generally be sufficient, and 1 inch 
in 30 inches should never be exceeded. 

Fig. 15. 

Side Ditches are provided to carry away the subsoil water 
from the base of the road, and the rain water which falls upon its 
surface; to do this speedily they must have capacity and inclination 



proportionate to the amount of water reaching them. The width 
of the bed should not be less than 18 inches; the depth will vary with 
circumstances, but should be such that the water surface shall not 
reach the subgrade, but remain at least 12 inches below the crown 
of the road. The sides should slope at least 1 to 1. 

The longitudinal inclination of the ditch follows the configura- 
tion of the general topography, that is, the lines of natural drainage. 
When the latter has to be aided artificially, grades from 1 in 500 to 
1 in 800 will usually answer. 

Fig. 10. 

In absorbent soil less fall is sufficient, and in certain cases level 
ditches are permissible. The slopes of the ditches must be protected 
where the grade is considerable. This can be accomplished by sod 
revetments, riprapping, or paving. 

These ditches may be placed either on the road or land side of the 
fence. In localities where open ditches are undesirable they may be 
constructed as shown in Figs. 13 to 17, and may be formed of stone 

Fig. 17. 

or tile pipe, according to the availability of either material. If for 
any reason two can not be built, build one. 

Springs found in the roadbed should be tapped and led into the 
side ditches. 

Drainage of the Surface. The drainage of the roadway 
surface depends upon the preservation of the cross-section, with 
regular and uninterrupted fall to the sides, without hollows or ruts 
in which the water can lie, and also upon the longitudinal fall of the 



road. If this is not sufficient the road becomes flooded during heavy 
rainstorms and melting snow, and is considerably damaged. 

The removal of surface water from country roads may be effected 
by the side ditches, into which, when there are no sidewalks, the 
water flows directly. When there are sidewalks, gutters are formed 
between the roadway and footpath, as shown in Figs. 13 to 17, and 
the water is conducted from these gutters into the side ditches by 
tile pipes laid under the walks at intervals of about 50 feet. The 
entrance to these pipes should be protected against washing by a 
rough stone paving. In the case of 'covered ditches under the footpath 
the water must be led into them by first passing through a catch 
basin. These are small masonry vaults covered with iron gratings to 
prevent the ingress of stones, leaves, etc. Connection from the 
catch basin is made by a tile pipe about 6 inches in diameter. The 
mouth of this pipe is placed a few feet above the bottom of the catch 
basin, and the space below it acts as a depository for the silt carried 
by the water, and is cleaned out periodically. The catch basins may 
be placed from 200 to 300 feet apart. They should be made of 
dimensions sufficient to convey the amount of water which is liable 
to flow into them during heavy and continuous rains. 

If on inclines the velocity of the water is greater than the nature 
of the soil will withstand, the gutters will be roughly paved. In all 
cases, the slope adjoining the footpath should be covered with sod. 

A velocity of 30 feet a minute will not disturb clay with sand and 
stone. 40 feet per minute will move coarse sand. 60 feet a minute 
will move gravel. 120 feet a minute should move round pebbles 1 inch 
in diameter, and 180 feet a minute will move angular stones If inches 
in diameter. 

The scour in the gutters on inclines may be prevented by small 
weirs of stones or fascines constructed by the roadmen at a nominal 
cost. At junctions and crossroads the gutters and side ditches re- 
quire careful arrangement so that the water from one road may not 
l>e thrown upon another; cross drains and culverts will be retjuired 
at such places. 

Water Breaks to turn the surface drainage into the side ditches 
should not be constructed on improved roads. They increase the 
grade and are an impediment to convenient and easy travel. Where 



it is necessary that water should cross the road a culvert should be 

On the side hill or mountain roads catch-water ditches should 
be cut on the mountain side above the road, to cut off and convey the 
drainage of the ground above them to the neighboring ravines. The 
size of these ditches will be determined by the amount of rainfall, 
extent of drainage from the mountain w r hich they intercept, and by the 
distances of the ravine water courses on each side. 

The inner road gutter should be of ample dimensions to carry 
off the water reaching it; when in soil, it should be roughly paved with 
stone. When paving is not absolutely necessary, but it is desirable 
to arrest the scouring action of running water during heavy rains, 
stone weirs may be erected across the gutter at convenient intervals. 
The outer gutter need not be more than 12 inches wide and 9 inches 
deep. The gutter is formed by a depression in the surface of the 
road close to the parapet or revetted earthen protection mound. The 
drainage which falls into this gutter is led off through the parapet, 
or other roadside protection at frequent intervals. The guard stones 
on the outside of the road are placed in and across this gutter, just 
below the drainage holes, so as to turn the current of the drainage 
into these holes or channels. On straight reaches, with parapet 
protection, drainage holes with guard stones should be placed every 
20 feet apart. Where earthen mounds are used and it may not be 
convenient to have the drainage holes or channels every 20 feet, the 
guardstones are to be placed in advance of the gutter to allow the 
drainage to pass behind them. This drainage is either to be run off 
at the cross drainage of the road, or to be turned off as before by a 
guard stone set across the gutter. 

At re-entering turns, where the outer side of the road requires 
particular protection, guard stones should be placed every 4 feet. 
As all re-entering turns should be protected by parapets, the drainage 
holes through them may be placed as close together as desired. 

Culverts are necessary for carrying under a road the streams 
it crosses, and also for conveying the surface water collected in the 
side ditches from the upper side to that side on which the natural 
water courses lie. 

Especial care is required to provide an ample way for the water 



to be passed. If the culvert is too small, it is liable to cause a washout, 
entailing interruption of traffic and cost of repairs, and possibly may 
cause accidents that will require payment of large sums for damages. 
On the other hand, if the culvert is made unnecessarily large, the 
cost of construction is needlessly increased. 

The area of waterway required depends (1) upon the rate of 
rainfall; (2) the kind and condition of the soil; (3) the character 
and inclination of the surface; (4) the condition and inclination of 
the bed of the stream; (5) the shape of the area to be drained, and 
the position of the branches of the stream ; (6) the form of the mouth 
and the inclination of the bed of the culvert; and (7) whether it is 
permissible to back the water up above the culvert, thereby causing 
it to discharge under a head. 

(1) It is the maximum rate of rainfall during the severest storms 
which is required in this connection. This varies greatly in different 
sections of the country. 

The maximum rainfall as shown by statistics is about one inch 
per hour (except during heavy storms), equal to 3,630 cubic feet per 
acre. Owing to various causes, not more than 50 to 75 per cent of 
this amount will reach the culvert within the same hour. 

Inches of rainfall X 3,630 = cubic feet per acre. 

Inches of rainfall X 2,323,200 = cubic feet per square mile. 

(2) The amount of water to be drained off will depend upon the 
permeability of the surface of the ground, which will vary greatly 
with the kind of soil, the degree of saturation, the condition of the 
cultivation, the amount of vegetation, etc. 

(3) The rapidity with which the water will reach the water 
course depends upon whether the surface is rough or smooth, steep 
or flat, barren or covered with vegetation, etc. 

(4) The rapidity with which the water will reach the culvert 
depends upon whether there is a well-defined and unobstructed 
channel, or whether the water finds its way in a broad thin sheet. 
It the water course is unobstructed and has a considerable inclination, 
the water may arrive at the culvert nearly as rapidly as it falls; but 
if the channel is obstructed, the water may be much longer in passing 
the culvert than in falling. 

(5) The area of waterway depends upon the amount of the area 



to be drained ; but in many cases the shape of this area and the posi- 
tion of the branches of the stream are of more importance than the 
amount of the territory. For example, if the area is long and narrow, 
the water from the lower portion may pass through the culvert before 
that from the upper end arrives; or, on the other hand, if the upper 
end of the area is steeper than the lower, the water from the former 
may arrive simultaneously with that from the latter. Again, if the 
lower part of the area is better supplied with branches than the upper 
portion, the water from the former will be carried past the culvert 
before the arrival of that from the latter; or, on the other hand, if 
the upper part is better supplied with branch water courses than 
the lower, the water from the whole area may arrive at the culvert 
at nearly the same time. In large areas the shape of the area and 
the position of the water courses are very important considerations. 

(6) The efficiency of a culvert may be very materially increased 
by so arranging the upper end that the water may enter into it without 
being retarded. The discharging capacity of a culvert can be greatly 
increased by increasing the inclination of its bed, provided the channel 
below will allow the water to flow away freely after having passed 
the culvert. 

(7) The discharging capacity of a culvert can be greatly increased 
by allowing the water to dam up above it. A culvert will discharge 
twice as much under a head of four feet as under a head of one foot. 
This can be done safely only with a well constructed culvert. 

The determination of the values of the different factors entering 
into the problem is almost wholly a matter of judgment. An estimate 
for any one of the above factors is liable to be in error from 100 to 
200 per cent, or even more, and of course any result deduced from 
such data must be very uncertain. Fortunately, mathematical exact- 
ness is not required by the problem nor warranted by the data. The 
question is not one of 10 or 20 per cent of increase; for if a 2-foot pipe 
is sufficient, a 3-foot pipe will probably be the next size, an increase 
of 225 per cent; and if a G-foot arch culvert is too small, an 8-foot will 
be used, an increase of ISO per cent. The real question is whether 
a 2-foot pipe or an S-foot arch culvert is needed. 

Valuable data on the proper size of any particular culvert may 
be obtained (1) by observing the existing openings on the same 


stream; (2) by measuring, preferably at time of high water, a cross- 
section of the stream at some narrow place; and (3) determining the 
height of high water as indicated by drift and the evidence of the 
inhabitants of the neighborhood. 

On mountain roads or roads subjected to heavy rainfall culverts 
of ample dimensions should be provided wherever required, and it 
will be more economical to construct them of masonry. In localities 
where boulders and other debris are likely to be washed down during 
wet weather, it will be a good precaution to construct catch pools at 
the entrance of all culverts and cross drains for the reception of 
such matter. In hard soil or rock these catch pools will be simple 
\vell-like excavations, with their bottom two or three feet below the 
entrance sill or floor of the culvert or drain. Where the soil is soft 
they should be lined with stone laid dry; if very soft, with masonry. 
The size of the catch pools will depend upon the width of the drainage 
works. They should be wide enough to prevent the drains from 
being injured by falling rocks and stones of a not inordinate size. 

The use of catch pools obviates the necessity of building culverts 
and drains at an angle to the axis of the road. Oblique structures 
are objectionable, as being longer than if set at. right angles and by 
reason of the acute- and obtuse-angled terminations to their piers, 
abutments, and coverings. 

Materials for Culverts. Culverts may be of stone, brick, vitri- 
fied earthenware, or iron pipe. Wood should be absolutely avoided. 

For small streams and a limited surface of rainfall either class 
of pipes, in sizes varying from 12 to 24 inches in diameter, will serve 
excellently. They are easily laid, and if properly bedded, with the 
earth tamped about them, are very permanent. Their upper surface 
should be at least 18 inches below the road surface, and the upper 
end should be protected with stone paving so arranged that the water 
can in no case work in around the pipe. 

When the flow of water is estimated to be too great for two lines 
of 24-inch pipes, a culvert is required. If stone abounds, it may be 
built of large roughly squared stones laid either dry or in mortar. 
When the span required is more than 5 feet, arch culverts either of 
stone or brick masonry may be employed. For spans above 15 feet 
the structure required becomes a bridge 




Earthenware Pipe Culverts. Construction. In laying the 
pipe the bottom of the trench should be rounded out to fit the lower 
half of the body of the pipe with proper depressions for the sockets. 
If the ground is soft or sandy, the earth should be rammed carefully, 
but solidly in and around the lower part of the pipe. The top surface 
of the pipe should, as a rule, never be less than 18 inches below the 
surface of the roadway, but there are many cases where pipes have 
stood for several years under heavy loads with only 8 to 12 inches of 
earth over them. No danger from frost need be apprehended, pro- 
vided the culverts are so constructed that the water is carried away 
from the level end. Ordinary soft drain tiles are not in the least 
affected by the expansion of frost in the earth around them. 

The freezing of water in the pipe, particularly if more than half 
full, is liable to burst it; consequently the pipe should have a suffi- 
cient fall to drain itself, and the outside should be so low that there 
is no danger of back waters reaching the pipe. If properly drained, 
there is no danger from frost. 

Jointing. In many cases, perhaps in most, the joints are 
not calked. If this is not done, there is liability of the water being 
forced out of the joints and washing away the soil from around the 
pipe. Even if the danger is not very imminent, the joints of the 
larger pipes, at least, should be calked with hydraulic cement, since 
the cost is very small compared with the insurance against damage 

Fig. 18. 

thereby secured. Sometimes the joints are calked with clay. Every 
culvert should be built so that it can discharge water under a head 
without damage to itself. 



Although often omitted, the end sections should be protected 
with a masonry or timber bulkhead. The foundation of the bulk- 
head should be deep enough not to be disturbed by frost. In con- 
structing the end wall, U is well to increase the fall near the outlet 
to allow for a possible settlement of the interior sections. When 
stone and brick abutments are too expensive, a fair substitute can 
be made by setting posts in the ground and spiking plank to them. 
When planks are used, it is best to set them with considerable inclina- 
tion towards the roadbed to prevent their being crowded outward 
by the pressure of the embankment. The upper end of the culvert 
should be so protected that the water will not readily find its way 

along the outside of the pipes, in case the mouth of the culvert should 
become submerged. 

When the capacity of one pipe is not sufficient, two or more 
may be laid side by side as shown in Fig. 19. Although the two 
small pipes do not have as much discharging capacity as a single 
large one of equal cross-section, yet there is an advantage in laying 
two small ones side by side, since the water need not rise so high 
to utilize the full capacity of the two pipes as would be necessary 
to discharge itself through a single one of large size. 

Iron Pipe Culverts. During recent years iron pipe has been 
used for culverts on many prominent railroads, and may be used on 
roads in sections where other materials are unavailable. 

In constructing a culvert with cast-iron pipe the points requiring 


particular attention are (1) tamping the soil tightly around the pipe 
to prevent the water from forming a channel along the outside, and 
(2) protecting the ends by suitable head walls and, when necessary, 
laying riprap at the lower end. The amount of masonry required 
for the end walls depends upon the relative width of the embankment 
and the number of sections of pipe used. For example, if the em- 
bankment is, say, 40 feet wide at the base, the culvert may consist of 
three 12-foot lengths of pipe and a light end wall near the toe of 
the bank; but if the embankment is, say, 32 feet wide, the culvert 
may consist of two 12-foot lengths of pipe and a comparatively heavy 
end wall well back from the toe of the bank. The smaller sizes of 
pipe usually come in 12- foot lengths, but sometimes a few 6-foot 

Broken Stones 
or Bricks 

Fig. 20. Section of Pipe Culvert 

lengths are included for use in adjusting the length of the culvert, 
to the width of the bank. The larger sizes are generally 6 feet long. 


The term "earthwork" is applied to all the operations per- 
formed in the making of excavation and embankments. In its 
widest sense it comprehends work in rock as well as in the looser 
materials of the earth's crust. 

Balancing Cuts and Fills. In the construction of new roads, 
the formation of the roadbed consists in bringing the surface of the 
ground to the adopted grade This grade should be established so as 



to reduce the earthwork to the least possible amount, both to render 
the cost of construction low, and to avoid unnecessary marring the 
appearance of the country in the vicinity of the road. The most 
desirable position of the grade line is usually that which makes the 
amount of cutting and filling equal to each other, for any surplus 
embankment over cutting must be made up by borrowing, and surplus 
cutting must be wasted, both of these operations involving additional 
cost for labor and land. 

Inclination of Side Slopes. The proper inclination for the 
side slopes of cutting and embankments depends upon the nature of 
the soil, the action of the atmosphere and of internal moisture upon 
it. For economy the inclination should be as steep as the nature 
of the soil will permit. 

The usual slopes in cuttings are: 

Solid rock . 1 to 1 

Earth and Gravel 3^ to 1 

Clay .3 or G to 1 

Fine sand 2 or 3 to 1 

The slopes of embankment are usually made 1 -\- to 1 . 
Form of Side Slopes. The natural, strongest, and ultimate 
form of earth slopes is a concave curve, iii which the flattest portion 
is at the bottom. This form is very rarely given to the slopes in con- 
structing them; in fact, the reverse is often the case, the slopes being 
made convex, thus saving excavation by the contractor and inviting 

In cuttings exceeding 10 feet in depth the forming of concave 
lopes will materially aid in preventing slips, and in any case they will 

Fig. 21 . Cross-Section for Embankment. 

reduce the amount of material which will eventually have to be re- 
moved when cleaning up. Straight or convex slopes will continue 
to slip until the natural form is attained. 

A revetment or retaining wall at the base of a slope will save 



In excavations of considerable depth, and particularly in soils 
liable to slips, the slope may be formed in terraces, the horizontal 
offsets or benches being made a few feet in width with a ditch on 
the inner side to receive the surface water from the portion of the 
side slope above them. These benches catch and retain earth 
that may fall from the slopes above them. The correct forms for ths 
slopes of embankment and excavation are shown in Figs. 21 and 22. 

Covering of Slopes. It is not usual to employ any artificial 
means to protect the surface of the side slopes from the action of the 
weather; but it is a precaution which in the end will save much labor 

Fig. 22. Cross-Section for Excavation. 

and expense in keeping the roadways in good order. The simplest 
means which can be used for this purpose consists in covering the 
slopes with good sods, or else with a layer of vegetable mould about 
four inches thick, carefully laid and sown with grass seed. These 
means are amply sufficient to protect the side slopes from injury 
when they are not exposed to any other cause of deterioration than 
the wash of the rain and the action of frost on the ordinary moisture 
retained by the soil. 

A covering of brushwood or a thatch of straw may also be used 
with good effect; but from their perishable nature they will require 
frequent renewal and repairs. 

Where stone is abundant a small wall of stone laid dry may be 
constructed at the foot of the slopes to prevent any wash from them 
being carried into the ditches. 

Shrinkage of Earthwork. All materials when excavated 
increase in bulk, but after being deposited in banks subside or shrink 
(rock excepted) until they occupy less space than in the pit from 
which excavated. 

Rock, on the other hand, increases in volume by being broken 
up, and does not settle again into less than its original bulk. The 
.increase may be taken at 50 per cent, 



The shrinkage in the different materials is about as follows: 

.Gravel 8 per cent 

Gravel and sand 9 " " 

Clay and clay earths 10 " ' 

Loam and light sandy earths 12 " " 

Loose vegetable soil. . . 15 " " 

Puddled clay 25 " 

Thus an excavation of loam measuring 1,000 cubic yards will 
form only about 880 cubic yards of embankment, or an embankment 
of 1,000 cubic yards will require about 1,120 cubic yards measured 
in excavation to make it. A rock excavation measuring 1,000 yards 
will make from 1,500 to 1,700 cubic yards of embankment, depending 
upon the size of the fragments. 

The lineal settlement of earth embankments will be about in 
the ratio given above; therefore either the contractor should be 
instructed in setting his poles to guide him as to the height of grade 
on an earth embankment to add the required percentage to the fill 
marked on the stakes, or the percentage may be included in the 
fill marked on the stakes. In rock embankments this is not necessary. 

Classification of Earthwork. Excavation is usually classi- 
fied under the heads earth, hardpan, loose rock, and solid rock. For 
each of these classes a specific price is usually agreed upon, and an 
extra allowance is sometimes made when the haul or distance to 
which the excavated material is moved exceeds a given amount. 

The characteristics which determine the classes to which a given 
material belongs are usually described with clearness in the speci- 
fications, as: 

Earth will include loam, clay, sand, and loose gravel. 

Hardpan will include cemented gravel, slate, cobbles, and boul- 
ders containing less than one cubic foot, and all other matters of an 
earthy nature, however compact they may be. 

Loose rocfc will include shale, decomposed rock, boulders, and 
detached masses of rock containing not less than three cubic feet, 
and all other matters of a rock nature which may be loosened with a 
pic-k, although blasting may be resorted to in order to expedite the 
work. ~ 

Solid rock will include all rock found in place in ledges and 



masses, or boulders measuring more than three cubic feet, and which 
can only be removed by blasting. 

Prosecution of Earthwork. No general rule can be laid 
down for the exact method of carrying on an excavation and dis- 
posing of the excavated material. The operation in each case can 
only be determined by the requirements of the contract, character 
of the material, magnitude of the work, length of haul, etc. 

Formation of Embankments. Where embankments are to be 
formed of less than two feet in height, all stumps, weeds, etc. should 
be removed from the space to be occupied by the embankment. 
For embankments exceeding two feet in height stumps need only 
be close cut. Weeds and brush, however, ought to be removed and 
if the surface is covered with grass sod, it is advisable to plow a fur- 
row at the toe of the slope. Where a cutting passes into a fill all 
the vegetable matter should be removed from the surface before 
placing the fill. The site of the bank should be carefully examined 
and all deposits of soft, compressible matter removed. WTien a bank 
is to be made over a swamp or marsh, the site should be thoroughly 
drained, and if possible the fill should be started on hard bottom. 

Perfect stability is the object aimed at, and all precautions neces- 
sary to this end should be taken. Embankments should be built in 
successive layers, banks two feet and under in layers from six 
inches to one foot, heavier banks in layers 2 and 3 feet thick. The 
horses and vehicles conveying the materials should be required to 
pass over the bank for the purpose of consolidating it, and care 
should be taken to have the layers dip towards the center. Embank- 
ments first built up in the center, and afterwards widened by dump- 
ing the earth over the sides, should not be allowed. 

Embankments on Hillsides. When the axis of the road 
is laid out on the side slope of a hill, and the road is formed partly 
by excavating and partly by embanking, the usual and most simple 
method is to extend out the embankment gradually along the whole 
line of the excavation. This method is insecure; the excavated 
material if simply deposited on the natural slope is liable to slip, 
and no pains should be spared to give it a secure hold, particularly 
at the toe of the slope. The natural surface of the slope should be 
cut into steps as shown in Figs. 23 and 24. The dotted line AB 



By means of the truck elevators here shown, the roadwav is carried over the steep bluffs of the Pal- 
isades of the Hudson at Hoboken, N. J. 

Copyright, 1907, by Underwood & Underwood, New York 



represents the natural surface of the ground, C E B the excavation, 
and ADC the embankment, resting on steps which have been cut 
between A and C. The best position for these steps is perpendicular 
to the axis of greatest pressure. If A D is inclined at the angle of 
repose of the material, the steps near A should be inclined in the 

Fig. 23. Method of Construction on Hillsides. 

opposite direction to A D, and at an angle of nearly 90 degrees 
thereto, while the steps near C may be level. If stone is abundant, 
the toe of the slope may be further secured by a dry wall of stone. 
On hillsides of great inclination the above method of construc- 
tion will not be sufficiently secure; retaining walls of stone must 
be substituted for the side slopes of both the excavations and em- 
bankments. These walls may be made of stone laid dry, when stone 

Fig. 24. Hillside Road with Retaining and Revetment Walls. 

can be procured in blocks of sufficient size to render this kind of con- 
struction of sufficient stability to resist the pressure of the earth. 
When the stones laid dry do not offer this security, they must be laid 
in mortar. The wall which forms the slope of the excavation should 
be carried up as high as the natural surface of the ground. Unless 
the material is such that the slope may be safely formed into steps 
or benches as shown in Fig. 23, the wall that sustains the embank- 
ment should l)e built up to the surface of the roadway, and a parapet 



wall or fence raised upon it, to protect pedestrians against accident. 
(See Fig. 24.) 

For the formula for calculating the dimensions of retaining walls 
see instruction paper on Masonry Construction. 

Roadways on Rock Slopes. On rock slopes when the in- 
clination of the natural surface is not greater than one perpendicular 
to two base, the road may be constructed partly in excavation and 
partly in embankment in the usual manner, or by cutting the face 
of the slope into horizontal steps with vertical faces, and building 
up the embankment in the form of a solid stone wall in horizontal 
courses, laid either dry or in mortar. Care is required in proportion- 
ing the steps, as all attempts to lessen the quantity of excavation by 
increasing the number and diminishing the width of the steps require 
additional precautions against settlement in the built-up portion 
of the roadway. 

When the rock slope has a greater inclination than 1 : 2 the 
whole of the roadway should be in excavation. 

In some localities roads have been constructed along the face 
of nearly perpendicular cliffs on timber frameworks consisting of 
horizontal beams, firmly fixed at one end by being let into holes 
drilled in the rock, the other end being supported by an inclined 
strut resting against the rock in a shoulder cut to receive it. There 
are also examples of similar platforms suspended instead of being 

Earth Roads. The term "earth road" is applied to roads 
where the surface consists of the native soil; this class of road is the 
most common and cheapest in first cost. At certain seasons of the 
year earth roads when properly cared for are second to none, but 
during the spring and wet seasons they are very deficient in- the im- 
portant requisite of hardness, and are almost impassable. 

For the construction of new earth roads, all the principles pre- 
viously discussed relating to alignment, grades, drainage, width, etc., 
should be carefully, followed. The crown or transverse contour 
should be greater than in stone roads. Twelve inches at the center 
in 25 feet will be sufficient. 

Drainage is especially important, because the material of the 
road is more susceptible to the action of water, and more easily 




destroyed by it than are the materials used in the construction of the 
better class of roads. When water is allowed to stand upon the 
road, the earth is softened, the wagon wheels penetrate it and the 
horses' feet mix and kneed it until it becomes impassable mud. The 
action of frost is also apt to be more disastrous upon the more per- 
meable surface of the earth road, having the effect of swelling and 
heaving the roadway and throwing its surface out of shape. It may 

Fig. 25. Bush Hooks. 

in fact be said that the whole problem of the improvement and 
maintenance of ordinary country roads is one of drainage. 

In the preparation of the wheelway all stumps, brush, vegetable 
matter, rocks and boulders should be removed from the surface and 
the resulting holes filled in with clean earth. The roadbed having 

Fig. 26. Axe Mattock. 

Fig. 27. Bush Mattock. 

been brought to the required grade and crown should be thoroughly 
rolled, all inequalities appearing during the rolling should be filled 
up and re-rolled. 

Care of Earth Roads. If the surface of the roadway is prop- 
erly formed and kept smooth, the water will be shed into the side 
ditches and do comparatively little harm; but if it remains upon the 
surface, it will be absorbed and convert the road into mud. All 
ruts and depressions should be filled up as soon as they appear. 
Repairs should be attended to particularly in the spring. At this 
season a judicious use of a road machine and rollers will make a 



smooth road. In summer when the surface gets roughed up it can 
be improved by running a harrow over it; if the surface is a little 
muddy this treatment will hasten the drying. 

During the fall the surface should be repaired, with special 
reference to putting it in shape to withstand the ravages of winter. 
Saucer-like depressions and ruts should be filled up with clean earth 
similar to that of the roadbed and tamped into place. 

The side ditches should be examined in the fall to see that they 
are free from dead weeds and grass, and late in winter they should 
be examined again to see that they are not clogged. The mouths of 
culverts should be cleaned of rubbish and the outlet of tile drains 
opened. Attention to the side ditches will prevent overflow, and 
washing of the roadway, and will also prevent the formation of ponds 
at the roadside and the consequent saturation of the roadbed. 

Holes and ruts should not be filled with stone, bricks, gravel 
or other material harder than the earth of the roadway as the hard 
material will not wear uniform with the rest of the road, but produce 
bumps and ridges, and usually result in making two holes, each 
larger than the original one. It is bad practice to cut a gutter from 
a hole to drain it to the side of the road. Filling is the proper course, 
whether the hole is dry or contains mud. 

In the maintenance of clay roads neither sods nor turf should 
be used to fill holes or ruts; for, though at first deceptively tough, 
they soon decay and form the softest mud. Neither should the ruts 
be filled with field stones; they will not wear uniformly with the rest 
of the road, but will produce hard ridges. 

Trees and close hedges should not be allowed within 200 feet 
of a clay road. It requires all the sun and wind possible to keep its 
surface in a dry and hard condition. 

Sand Roads. The aim in the improvement of sand roads is to 
have the wheelway as narrow and well defined as possible, so as to 
have all the vehicles run in the same track. An abundant growth 
of vegetation should be encouraged on each side of the wheelway, 
for by this means the shearing of the sand is, in a great measure, 
avoided. Ditching beyond a slight depth to carry away the rain 
water is not desirable, for it tends to hasten the drying of the sands,, 
which is to be avoided. Where possible the roads should be over- 



hung with trees, the leaves and twigs of which catching on the 
wheelway will serve still further to diminish the effect of the wheels 
in moving the sands about. If clay can be obtained, a coating 6 
inches thick will be found a most effective and economical improve- 
ment. A coating of 4 inches of loose straw will, after a few days' 
travel, grind into the sand and become as hard and as firm as a 
dry clay road. 

The maintaining of smooth surfaces on all classes of earth roads 
will be greatly assisted and cheapened by the frequent use of a roller 
(either steam or horse) and any one of the various forms of road 
grading and scraping machines. In repairing an earth road the 
plough should not be used. It breaks up the surface which has 
been compacted by time and travel. 


Picks are made of various styles, according to the class of 
material in which they are to be used. Fig. 28 shows the form 

Fig. 29. Clay Pick. 

usually employed in street work. Fig. 29 shows the form generally 
used for clay or gravel excavation. 

The eye of the pick is generally formed of wrought iron, pointed 
with steel. The weight of picks ranges from 4 to 9 Ib. 

Pig. 30. Shovels. 

Shovels are made in two forms, square and round pointed, 
usually of pressed steel. 

Ploughs are extensively employed in grading, special forms 
being manufactured for the purpose. They are known as " grading 
ploughs," " road ploughs" " township ploughs," etc. They vary 



in form according to the kind of work they are intended for, viz. : 
loosening earth, gravel, hardpan, and some of the softer rocks. 

These ploughs are made of great strength, selected white oak, 
rock elm, wrought steel and iron being generally used in their con- 
struction. The cost of operating ploughs ranges from 2 to 5 cents 
per cubic yard, depending upon the compactness of the soil. The 
quantity of material loosened will vary from 2 to 5 cubic yards per 

Fig. 31 shows the form usually adopted for loosening earth. 
This plough does not turn the soil, but cuts a furrow about 10 

Fig. 31. Grading Plow. 

inches wide and of a depth adjustable up to 11 inches. 

In light soil the ploughs are operated by two or four horses; in 
heavy soils as many as eight are employed. Grading ploughs vary 
in weight from 100 to 325 Ib. 

Pig. 32. Hardpan Plow. 

Fig. 32 illustrates a plough specially designed for tearing up 
macadam, gravel, or similar material. The point is a straight bar 
of cast steel drawn down to a point, and can be easily repaired. 




Scrapers are generally used to move the material loosened by 
ploughing; they are made of either iron or steel, and in a variety 
of form, and are known by various names, as " drag," " buck," 
" pole," and " wheeled". The drag scrapers are usually employed 
on short hauls, the wheeled on long hauls. Fig. 33 illustrates the 
usual form of drag scrapers. 

Drag scrapers are made in three sizes. The smallest, for one 
horse, has a capacity of 3 cubic feet; the others, for two horses, 

Fig. 33. Drag Scraper. 

have a capacity of 5 to 7- cubic feet. The smallest weighs about 
90 lb., and the larger ones from 94 to 102 Ib. 

Buck scrapers are made in two sizes two-horses, carrying 1\ 
cubic feet; four-horses, 12 cubic feet. 

Pole scraper, Fig. 34, is designed for use in making and leveling 
earth roads and for cutting and cleaning ditches; it is also well 

Fig. 34. Pole Scraper. 

adapted for moving earth short distances at a minimum cost. 

Wheeled scrapers consist of a metal box, usually steel, mounted 
~on wheels, and furnished with levers for raising, lowering, and 




dumping. They are operated in the same manner as drag scrapers, 
except that all the movements are made by means of the levers, and 
without stopping the team. By their use the excessive resistance to 

Fig. 35. Wheeled Scraper. 

traction of the drag scraper is avoided. Various sizes are made, 
ranging in capacity from 10 to 17 cubic feet. In weight they range 
from 350 to 700 Ib. 

Wheelbarrows. The wheelbarrow shown in Fig. 36 is con- 
structed of wood and is the most commonly employed for earth- 
work. Its capacity ranges from 2 to 2J cubic feet. Weight about 
50 Ib. 

The barrow, Fig. 37, has a pressed-steel tray, oak frame, and 
steel wheel, and will be found more durable in the maintenance 

Fig. 36. Wooden Barrow. 

department than the all wood barrow. Capacity from 3^ to 5 cubic 
feet, depending on size of tray. 

The barrow, Fig. 38, is constructed with tubular iron frames 
and steel tray, and is adaptable to the heaviest work, such as 




moving heavy broken' stone, etc., or it may be employed with ad- 
vantage in the cleaning department. Capacity from 3 to 4 cubic 
feet. Weight from 70 to 82 Ib. 

Fig. 37. Steel Tray Barrow. 

The maximum distance to which earth can be moved economic- 
ally in barrows is about 200 feet. The wheeling should be per- 
formed upon planks, whose steepest inclination should not exceed 1 
in 12. The force required to move a barrow on a plank is about oV 
part of the weight; on hard dry earth, about y part of the weight. 

Fig. 38. Metal Barrow. 

The time occupied in loading a barrow will vary with the 
character of the material and the proportion of wheelers to shovel- 
lers. Approximately, a shoveller takes about as long to fill a barrow 
with earth as a wheeler takes to wheel a full barrow a distance of 
about 100 or 120 feet on a horizontal plank and return with the 
empty barrow. 

Carts. The cart usually employed for hauling earth, etc., is 
shown in Fig. 39. The average capacity is 22 cubic feet, and the 
average weight is 800 Ib. These carts are usually furnished with 
broad tires, and the body is so balanced that the load is evenly 
divided about the axle. 



The time required to load a cart varies with the material. One 
shoveller will require about as follows: Clay, seven minutes; loam, 
six minutes; sand, five minutes. 

Fig. 39. Earth Wagon. 

Dump Cars. These cars are made to dump in several different 
ways, viz., single or double side, single or double end, and rotary 
or universal dumpers. 

Dump cars may be operated singly or in trains, as the magni- 
tude of the work mty demand. They may be moved by horses or 

Pig. 40. Dump Cart. 

small locomotives. They are made in various sizes, depending upon 
the gauge of the track on which they are run. A common gauge is 


20 inches, but it varies from that up to the standard railroad gauge 
of 662 inches. 

Dump Wagons. (Fig. 40.) The use of these wagons for mov- 
ing excavated earth, etc., and for transporting materials such as sand, 
gravel, etc., materially shortens the time required for unloading the 
ordinary form of contractor's wagon; having no reach or pole con- 
necting the rear axle with the center bearing of the front axle, they 
may be cramped short and the load deposited just where required. 
They are operated by the driver, and the capacity ranges from 35 
to 45 cubic feet. 

Mechanical Graders are used extensively in the making and 
maintaining of earth roads. They excavate and move earth more 
expeditiously and economically than can be done by hand; they are 
called by various names, such as "road machines," "graders," 
"road hones," etc. Their general form is shown in Fig. 41. 

Briefly described, they consist of a large blade made entirely 
of steel or of iron, or wood shod with steel, which is so arranged by 
mechanism attached to the frame from which it is suspended that it 
can be adjusted and fixed in any direction by the operator. In their 
action they combine the work of excavating and transporting the 

Fig. 41. Mechanical Grader. 

earth. They have been chiefly employed in the forming and main- 
tenance of earth roads, but may be also advantageously used in pre- 
paring the subgrade surface of roads for the reception of broken 
stone or other improved covering. 

A large variety of such machines are on the market. The 
"New Era" grader excavates the material from side ditches, and 



automatically loads the material into carts or wagons. Briefly de- 
scribed, the machine consists of a plough which loosens and raises 
the earth, depositing it upon a transverse carrying-belt, which con- 
veys it from excavation to embankment. This carrier is built in four 
sections, bolted together, so it can be used to deliver earth at 14, 
17, 19, or 22 feet from the plough. The carrier belt is of heavy 
3-ply rubber 3 feet wide. 

The plough and carrier are supported by a strong trussed frame- 
work resting on heavy steel axles and broad wheels. The large 
rear wheels are ratcheted upon the axle, and connected with strong 
gearing which propels the carrying-belt at- right angles to the direc- 
tion in which the machine is moving. 

The wheels and trusses are low and broad, occupying a space 
8 feet wide and 14 feet long, exclusive of the side carrier. This 
enables it to work on hillsides where any wheeled implements can 
be used. Notwithstanding its large size it is so flexible that it may 
be turned around on a 16-foot embankment. Pilot wheels and 
levers enable the operator to raise or lower the plough or carrier at 

. As a motive power 12 horses 8 driven in front, 4 abreast, and 
4 in the rear on a push cart are usually employed. 

When the teams are started, the operator lowers the plough and 
throws the belting into gear, and as the plough raises and turns the 
earth to the side the belt receives and delivers it at the distance for 
which the carrier is adjusted, forming either excavation or embank- 

When it becomes necessary to deliver the excavated earth beyond 
the capacity of the machine (22. feet or 1\ feet above the plough), 
the earth is loaded upon wagons, then conveyed to any distance. 
Arranging the carrier at 19 feet, wagons are. driven under the car- 
rier and loaded with \\ to H yards of earth in from 20 to 30 
seconds. When one wagon turns out with its load, another drives 
under the carrier, and the machine thus loads 600 to 800 wagons 
per day. It is claimed that with six teams and three men it is capa- 
ble of excavating and placing in embankment from 1000 to 1500 
cubic yards of earth in ten hours, or of loading from 600 to 800 



wagons in the same time, and that the cost of this handling is from 
1^ to 2^ cents per cubic yard. 

Points to be Considered in Selecting a Road Machine. In 

the selection of a road machine the following points should be care- 
fully considered: 

(1) Thoroughness and simplicity of its mechanical construction. 

(2) Material and workmanship used in its construction. 

(3) Ease of operation. 

(4) Lightness of draft 

(5) Adaptability for doing general road- work, ditching, etc. 

(6) Safety to the operator. 

Care of Road Machines. The road machine when not in use 
should be stored in a dry house and thoroughly cleaned, its blade 
brushed clean from all accumulations of mud, wiped thoroughly dry, 
and well covered with grease or crude oil. The axles, journals, and 
wearing parts should be kept well oiled when in use, and an extra 
blade should be kept on hand to avoid stopping tHe machine while 
the dulled one is being sharpened. 

Surface Graders. The surface grader, Fig. 42, is used for re- 
moving earth previously loosened by a plough. It is operated by 
one horse. The load may be retained and carried a considerable 

Fig. 42. Surface Grader. 

distance, or it may be spread gradually as the operator desires. It 
is also employed to level off and trim the surface after scrapers. 

The blade is of steel, J-inch thick, 15 inches wide, and 30 
inches long. The beam and other parts are of oak and iron. 
Weight about 60 Ib. 




The road leveller, Fig. 43, is used for trimming and smoothing 
the surface of earth roads. It is largely employed in the Spring 
when the frost leaves the ground. 

Fig. 43. Road Leveller. 

NO. 3, MO. 4. NO. 5. 


No. 6. 

No. /. No. 7. 

Fig. 44. Draining Tools. 

The blade is of steel, ^-inch thick by 4 inches by 72 inches, and 
is provided with a seat for the driver. It is operated by a team of 
horses. Weight about 150 Ib. 


Draining=tools. The tools employed for digging the ditches 
and shaping the bottom to fit the drain tiles are shown in Fig. 44. 
They are convenient to use, and expedite the work by avoiding 
unnecessary excavation. 

The tools are used as follows: Nos. 3, 4 and 5 are used for 
digging the ditches; Nos. 6 and 7 for cleaning and rounding the 

Fig. 45. Reversible Roller. 

bottom of the ditch for round tile. No. 2 is used for shoveling out 
loose earth and levelling the bottom of the ditch; No. 1 is used for 
the same purpose when the ditch is intended for "sole " tile. 

Fig. 46. Watering Cart. 

Horse Rollers. There is a variety of horse rollers on the 
market. Fig. 45 shows the general form. Each consists essentially 
of a hollow cast-iron cylinder 4 to 5 feet long, 5 to G feet in 


diameter, and weighing from 3 to 6 tons. Some forms are provided 
with boxes in which stone or iron may be placed to increase the 
weight, and some have closed ends and may be filled with water or 

Sprinkling-carts. Fig. 46 shows a convenient form of sprink- 
ling cart for suburban streets and country roads. Capacity about 
150 gallons. 


Road coverings consist of some foreign material as gravel, 
broken stone, clay, etc., placed on the surface of the earth road. 
The object of this covering, whatever its nature, is (1) to protect the 
natural soil from the effect of weather and travel, and (2) to furnish 
a smooth surface on which the resistance to traction will be reduced 
to the least possible amount, and over which vehicles may pass with 
safety and expedition at all seasons of the year. . Where an artificial 
covering is employed, the wheel loads coming upon its surface are 
distributed over a greater area of the roadbed than if the loads 
come directly upon the earth itself. The loads are not sustained by 
the covering as a rigid structure, but are transferred through it to 
the roadbed, which must support both the weight of the covering 
and the load coming upon it. 

Gravel Roads. Gravel is an accumulation of small more or 
less rounded stones which usually vary from the size of 'a small pea 
to a walnut. It is often intermixed with other substances, such as 
sand, clay, loam, etc., from each of which it derives a distinctive 
name. In selecting gravel for road purposes the chief quality to be 
sought for is the property of binding. 

Gravel in general is unserviceable for roadmaking. This is 
due mainly to the fact that the surface of the pebbles is smooth, so 
that they will not bind together in the manner of broken stone. 
There is also an absence of dust or other material to serve as a 
binder, and even if such binding material is furnished it is difficult 
to effectively hold the rounded and polished surface of the pebbles 

In certain deposits of gravel, particularly where the pebbly 
matter is to a greater or less extent composed of limestone, a con- 
siderable amount of iron oxide has been gathered in the mass. 



This effect is due to the tendency of water which contains iron to 
lay down that substance and to take lime in its place when the 
opportunity for so doing occurs. Such gravels are termed ferru- 
ginous. They are commonly found in a somewhat cemented state, 
and when broken up and placed upon roads they again cement, even 
more firmly than in the original state, often forming a roadway of 
very good quality. 

When no gravel but that found in rivers or on the seashore can 
be obtained, one-half of the stone should be broken and mixed with 
the other half; to the stone so mixed a small quantity of clay or 
loam, about one-eighth of the bulk of the gravel, must be added: 
an excess is injurious. Sand is unsuitable. It prevents packing in 
proportion to the amount added. 

Preparing the Gravel. Pit gravel usually contains too much 
earth, and should be screened before being used. Two sieves should 
be provided, one with meshes of one and one-half inches, so that all 
pebbles above that size may be rejected, the other with meshes of 
three quarters of an inch, and the material which passes through it 
should be thrown away. The expense of screening will be more 
than repaid by the superior condition of the road formed by the 
cleaned material, and by the diminution of labor in keeping it in 
order. The pebbles larger than one and a half inches may be 
broken to that size and mixed with clean material. 

Laying the Gravel. On the roadbed properly prepared a layer 
of the prepared gravel four inches thick is uniformly spread over the 
whole width, then compacted with a roller weighing not less than 
two tons, and having a length of not less than thirty inches. The 
rolling must be continued until the pebbles cease to rise or creep in 
front of the roller. The surface must be moistened by sprinkling 
in advance of the roller, but too much water must not be used. 
Successive layers follow, each being treated in the above described 
manner until the requisite depth and form has been attained. 

The gravel in the bottom layer must be no larger than that in 
the top layer; it must be uniformly mixed, large and small together, 
for if not, the vibration of the traffic and the action of frost will 
cause the larger pebbles to rise to the surface and the smaller ones 
to descend, and the road will never be smooth or firm. 



The pebbles in a gravel road are simply imbedded in a paste 
and can be easily displaced. It is for this reason, among others, 
that such roads are subject to internal destruction. 

The binding power of clay depends in a large measure upon 
the state of the weather. During rainy periods a gravel road be- 
comes soft and muddy, while in very dry weather the clay will con- 
tract and crack, thus releasing the pebbles, and giving a loose 
surface. The most favorable conditions are obtained in moderately 
damp or dry weather, during which a gravel road offers several 
advantages for light traffic, the character of the drainage, etc., 
largely determining durability, cost, maintenance, etc. 

Repair. Gravel roads constructed as above described will 
need but little repairs for some years, but daily attention is required 
to make these. A garden rake should be kept at hand to draw 
any loose gravel into the wheel tracks, and for filling any depres- 
sions that may occur. 

In making repairs, it is best to apply a small quantity of gravel 
at a time, unless it is a spot which has actually cut through. Two 
inches of gravel at once is more profitable than a larger amount. 
Where a thick coating is applied at once it does not all pack, and if, 
after the surface is solid, a cut be made, loose gravel will be found; 
this holds water and makes the road heave and become spouty 
under the action of frost. It will cost no more to apply six inches 
of gravel at three different times than to do it at once. 

At every one-eighth of a mile a few cubic yards of gravel 
should be stored to be used in filling depressions and ruts as fast as 
they appear, and there should be at least one laborer to every five 
miles of road. 

Broken Stone Roads. Broken stone roads are formed by pla- 
cing small angular fragments of stone on the surface of the earth 
roadbed and compacting into a solid mass by rolling. This class 
of road covering is generally called a Macadam or Telford road 
from the name of the two men who first introduced this type into 

The name of Telford is associated with a rough stone founda- 
tion, which he did not always use, but which closely resembled that 
which had been previously used in France. Macadam disregarded 


this foundation, contending that the subsoii, however bad, would 
carry any weight if made dry by drainage and kept dry by an im- 
pervious covering. The names of both have ever since been 
associated with the class of road which each favored, as well as with 
roads on which all their precepts have been disregarded. 

Quality of Stones. The materials used for broken-stone pave- 
ments must of necessity vary very much according to the locality. 
Owing to the cost of haulage, local stone must generally be used, 
especially if the traffic be only moderate. If, however, the traffic is 
heavy, it will sometimes be found better and more economical to 
obtain a superior material, even at a higher cost, than the local 
stone; and in cases where the traffic is very great, the best material 
that can be obtained is the most economical. 

The qualities required in a good road stone are hardness and 
and toughness and ability to resist the disintegrating action of the 
weather. These qualities are seldom found together in the same 
stone. Igneous and siliceous rocks, although frequently hard and 
tough, do not consolidate so well nor so quick as limestone, owing 
to the sandy detritus formed by the two first having no cohesion, 
whilst the limestone has a detritus which acts like mortar in binding 
the stones together. 

A stone of good binding nature will frequently wear much 
better than one without, although it is not so hard. A limestone 
road well made and of good cross-section will be more impervious 
than any other, owing to this cause, and will not disintegrate so 
soon in dry weather, owing partly to this and partly to the well- 
known quality which all limestone has of absorbing moisture from 
the atmosphere. Mere hardness without toughness is not of much 
use, as a stone may be very hard but so brittle as to be crushed to 
powder under a heavy load, while a stone not so hard but having a 
greater degree of toughness will be uninjured.' 

By a stone of good binding quality is meant one that, when 
moistened by water and subjected to the pressure of loaded wheels 
or rollers, will bind or cement together. This quality is possessed to 
a greater or less extent by nearly all rocks when in a state of dis- 
integration. The binding is caused by the action of water upon the 
chemical constituents of the stone contained in the detritus produced 


by crushing the stone, and by the friction of the fragments on each 
other while being compacted; its strength varies with the different 
species of rock, but it exists in some measure with them all, being 
greatest with limestone and least with gneiss. 

The essential condition of the stone to produce this binding 
effect is that it bj sound. No decayed stone retains the property of 
binding, though in some few cases, where the material contains iron 
oxides, it may, by the cementing property of the oxide, undergo a 
certain binding. 

A stone for a road surface should be as little absorptive of 
moisture as possible in order that it may not suffer injury from the 
action of frost. Many limestones are objectionable on this account. 

The stone used should be uniform in quality, otherwise it will 
wear unevenly, and depressions will appear where the softer material 
has been used. 

As the under parts of the road covering are not subject to the 
wear of traffic, and have only the weight of loads to sustain, it is 
not necessary that the stone of the lower layer be so hard or so 
tough as the stone for the surface, hence it is frequently possible by 
using an inferior stone for that portion of the work, to greatly reduce 
the cost of construction. 

Size of Stones. The stone should be broken into fragments 
as nearly cubical as possible. The size of the cubes will depend 
upon the character of the rock. If it be granite or trap, they should 
not exceed 1| inches in their greatest dimensions; if limestone, they 
should not exceed 2 inches. 

The smaller the stones the less the percentage of voids. Small 
stones compact sooner, require less binding, and make a smoother 
surface than large ones, but the size of the stone for any particular 
section of a road must be determined to a certain extent by the 
amount of traffic which it will have to bear and the character of the 
rock used. 

It is not necessary nor is it advisable that the stone should be 
all of the same size; they may be of all sizes under the maximum. 
In this condition the smaller stones fill the voids between the larger 
and less binding is required. 

Thickness of the Broken Stone. The offices of the broken 



stone are to endure friction and to shed water; its thickness must 
be regulated by the quality of the stone, the amount of traffic, and 
nature of the natural soil bed. Under heavy traffic it is advisable 
to make the thickness greater than for light traffic, in order to pro- 
vide for wear and lessen the cost of renewals. 

When the roadbed is firm, well drained, and not likely to be 
affected by ground water, it will always afford a firm foundation 
for the broken stone, the thickness of which may be made the mini- 
mum for good construction. This thickness is four inches. When 
this thickness is employed the stone must be of exceptionally fine- 
quality and the road must be maintained by the " continuous " 
method. With heavy traffic the thickness should be increased over 
the minimum a certain amount, say 2 inches, to provide for wear. 
Where the foundation is unstable and there is a tendency on the 
part of the loads to break through the covering, the thickness of 
the stone must be made the maximum, which is 12 inches. In such 
a case it may be advisable to employ a Telford foundation. Where 
the covering exceeds six inches in thickness, the excess may be 
composed of gravel, sand or ledge stone, the choice depending 
entirely on the cost, for all are equally effective. 

Foundation. The preparation of the natural soil over which 
the road is to be constructed, to enable it to sustain the superstruc- 
ture and the weights brought upon it, requires the observance of 
certain precautions the neglect of which will sooner or later result 
in the deterioration or possible destruction of the road covering. 
These precautions vary with the character of the soil. 

Soils of a siliceous and calcareous nature do not present any 
great difficulty, as their porous nature generally affords good natural 
drainage which secures a dry foundation. Their surface, however, 
requires to be compacted; this' is effected by rolling. 

The rolling should be carried out in dry weather, and any de- 
pressions caused by the passage of the roller should be filled with 
the same class of material as the surrounding soil. The rolling 
must be repeated until a uniform and solid bed is obtained. 

The argillaceous and allied soils, owing to their retentive 
nature, are very unstable under the action of water and frost, and 
in their natural condition afford a poor foundation. The prepara- 



tion of such soil is effected by drainage and by the application of a 
layer of suitable material to entirely separate the surface from the 
road material. This material may be sand, furnace ashes, or other 
material of a similar nature, spread in a layer from 3 to 6 inches 
thick over the surface of the natural soil. 

When the road is formed in rock cuttings it is advisable to 
spread a layer of sand or other material of light nature, so as to 
fill up the irregularities of the surface as well as to form a cushion 
for the road material to rest on. 

Spreading the Stone. The stone should be hauled upon the 
roadbed in broad-tire two-wheeled carts and dumped in heaps and 
be spread evenly with a rake in a layer which should be of a depth 
of 4J- inches. 

Watering. Wetting the stone expedites the consolidation, 
decreases crushing under the roller, and assists the filling of the 
voids with the binder. It should be applied by a sprinkler and 
should not be thrown on in quantity or from the plain nozzle of a 

Excessive watering, especially in the earlier stages, tends to 
soften the foundation, and care should be exercised in its appli- 

Binding. As the voids in loosely spread broken stone range 
from 35 to 50 per cent of the volume, and as no amount of rolling 
will reduce the voids more than one-half, it is necessary, in order to 
form an impervious and compact mass, to add some fine material 
which is called the binder. It may consist of the fragments and 
detritus obtained in crushing the stone. When this is insufficient, 
as will be. the case with the harder rocks, the deficiency may be 
made up of clean sand or gravel. The proportion of binder 
should slightly exceed the voids in the aggregate; it must not be 
mixed with the stones, but should be spread uniformly in small 
quantities over the surface and rolled into the interstices with the 
aid of water and brooms. 

As the quality of the binding used is of vital importance, the 
employment of inferior material, such as road scrapings or material 
of a clayey nature, should be avoided, even if the initial cost of the 
work should be greater when a good binding material is used. 



Stone consolidated with improper binding material may present 
a good appearance immediately after being rolled and be otherwise 
an apparently good piece of work, still in damp weather a consider- 
able amount of "lick-Ing up" by the wheels of the vehicles will take 
place, which reduces the strength of the coating and causes the sur- 
face to wear unequally. 

By the application of an immoderate quantity of binding of any 
description the stone coating will become unsound or rotten in con- 
dition, and if the binding be of an argillaceous nature, it will expand 
during frost, owing to its absorbent properties, and cause the dis- 
placement of the stones. The surface will become sticky, which 
seriously affects the tractive power of horses, while the road itself 
will suffer by the irregular deterioration of the surface. 

The use of such material as mentioned for binding enables 
rolling to be accomplished in much less time than when proper bind- 
ing is used, and the cost of consolidating the stone may be reduced 
by 25 per cent; but, on the other hand, the stone coating which will 
probably contain under these circumstances from 30 to 40 per cent 
of soft and soluble matter, and possibly present a smooth surface 
immediately after being rolled, will quickly become "cupped" by 
the wheel traffic, a bumpy surface being the result. This is caused 
by the irregular wear, while the lasting qualities or "life" of the 
coating will be shortened, giving unsatisfactory results to those 
traveling over the road, and the work of renewing the surface of 
the road in this manner may prove a failure on economical grounds. 
There can be no doubt, and it is now being more generally recog- 
nized, that sand as a material for binding in connection with rolling 
operations, when applied in a limited but sufficient quantity, pro- 
motes the durability- of the stone coating, while the general results 
are equally satisfactory; a firm, compact, and smooth surface is 
obtained, and the subsequent maintenance of the road is minimized. 

A great amount of rolling is necessary when sand is employed 
as a binding material, but economy is promoted, and the results are 
more satisfactory when sand is used than by the use of the material 
which gives to the stone an appearance only of having been properly 
consolidated. If clean sand be used in combination with the screen- 
ings from the crusher a very satisfactory surface will be obtained. 


If the use of motor vehicles equipped with pueumatic tires be- 
comes general, it is possible that some other description of binding 
material will be necessary. The pumping action of suction created 
by pneumatic tires, especially when propelled at a high speed, causes 
a considerable movement of the fine particles of the binding material, 
which on being displaced will convert the covering into a mass of 
stones. This objection can probably be overcome by watering. 

Compacting the Broken Stone. Three methods of compacting 
the broken stone are practiced : (1 ) by the traffic passing over the road ; 
(2) by rollers drawn by horses ; (3) by rollers propelled by steam. 

The first method is both defective and objectionable. (1) It is 
destructive to the horses and vehicles using the road. (2) It is waste- 
ful of material ; about one-third of the stone is worn away in the oper- 
ation. (3) Dung and dust are ground up with the stone, and the 
road is more readily affected by wet and frost. 

Steamrollers were first successfully introduced in France in 1860, 
since which time they have been almost universally adopted on account 
of the superiority and economy of the work done. Their use shortens 
the time required for construction or repair, and effects an indirect 
saving by the reduced wear and tear of horses and vehicles. They are 
made in different weights ranging from 3 to 30 tons. For the compact- 
ing of broken stone roads the weights in favor are from ten to fifteen 
tons; the heavier weights are considered unwieldy and their use is 
liable to cause damage to the underground structures that may be in 
the roadway. 

The advantage of steam rolling may be summed up as follows: 

(1) They shorten the time of construction. 

(2) A saving of road material, (a) because there are no loose 
stones to be kicked about and worn ; (b) because there is no abrasion 
of the stone, only one surface of the stone being exposed to wear; (c) 
because a thinner coating of stone can be employed; (d) because no 
ruts can be formed in which water can lie to rot the stone. 

(3) Steam-rolled roads are easier to travel on account of their 
even surface and superior hardness and they have a better appearance. 

(4) The roads can be repaired at any season of the year. 

(5) Saving both in materials and manual labor. 






The first work requiring the skill of the engineer is to lay out town 
sites properly, especially with reference to the future requirements of a 
large city where any such possibility exists. Few if any of our large 
cities were so planned. The same principles, to a limited extent, are 
applicable to all towns or cities. The topography of the site should be 
carefully studied, and the street lines adapted to it. These lines 
should be laid out systematically, with a view to convenience and 
comfort, and also with reference to economy of construction, future 
sanitary improvements, grades, and drainage. 

Arrangement of City Streets. Generally, the best method of 
laying out streets is in straight lines, with frequent and regular inter- 
secting streets, especially for the business parts of a city. When there 
is some centrally located structure, such as a courthouse, city hall, 
market, or other prominent building, it is very desirable to have several 
diagonal streets leading thereto. In the residence portions of cities, 
especially if on hilly ground, curves may with advantage replace 
straight lines, by affording better grades at less cost of grading, and by 
improving property through avoiding heavy embankments or cuttings. 

Width of Streets. The width of streets should be proportioned 
to the character of the traffic that will use them. No rule can be laid 
down by which to determine the best width of streets; but it may safely 
be said that a street which is likely to become a commercial thorough- 
fare should have a width of not less than 120 feet between the building 
lines the carriage-way 80 feet wide, and the sidewalks each 20 feet 

In streets occupied entirely by residences a carriage-way 32 feet 
wide will be ample, but the width between the building lines may be as 
great as desired. The sidewalks may be any amount over 10 feet 

Copyright, IMS, by American School of Corespondence. 




which fancy dictates. Whatever width is adopted for them, not more 
of it than 8 feet need be paved, the remainder being occupied with 
grass and trees. 

Street Grades. The grades of city streets depend upon the 
topography of the site. . The necessity of avoiding deep cuttings or 
high embankments which would seriously affect the value of adjoining 
property for building purposes, often demands steeper grades than 
are permissible on country roads. Many cities have paved streets 
on 20 per cent grades. In establishing grades through unimproved 
property, they may usually be laid with reference to securing the most 
desirable percentage within a proper limit of cost. But when improve- 
ments have already been made and have been located with reference 
to the natural surface of the ground, giving a desirable grade is fre- 
quently a matter of extreme difficulty without injury to adjoining 
property. In such cases it becomes a question of how far individual 

interests shall be sacrificed to the 
general good. There are, how- 
ever, certain conditions which it is 
important to bear in mind : 

(1) That the longitudinal 
crown level should be uniformly 
sustained from street' to street 
intersection, whenever practicable. 

(2) That the grade should 
be sufficient to drain the surface. 

(3) That the crown levels at 
all intersections should be ex- 
Fig. 47. tended transversely, to avoid form- 
ing a depression at the junction. 

Arrangements of Grades at Street Intersections, The best ar- 
rangement for intersections of streets when either or both have much 
inclination, is a matter requiring much consideration, and is one upon 
which much diversity of opinion exists. No hard or fast rule can be 
laid down; each will require special adjustment. The best and sim- 
plest method is to make the rectangular space aaaaaaaa, Fig. 47, 
level, with a rise of one-half inch in 10 feet from AAAA to B, placing 
gulleys at AAAA and the catch basins at ccc. When this method is 
not practicable, adopt such a grade (but one not exceeding 2| per cent) 




that the rectangle AAAA shall appear to he nearly level ; hut to secure 
this it must actually have a considerable dip in the direction of the 
slope of the street. If steep grades are continued 'across intersections, 
they introduce side slopes in the streets thus crossed, which are trouble- 
some, if not dangerous, to vehicles turning the corners, especially the 
upper ones. Such intersections are especially objectionable in rainy 

























I/.9B 31. 1 









(, N 

^ ff ?T 

Q h 



302 , 


i/^3 . S87e. 








Fig. 48. 

weather. The storm water will fall to the lowest point, concentrating 
a large quantity of water at two receiving basins, which, with a broken 
grade, could be divided between four or more basins. 

Fig. 48 shows the arrangement of intersections in steep grades 
adapted for the streets of Duluth, Minn. From this it will be seen 
that at these intersections the grades are flattened to three per cent for 
the width of the roadway of the intersecting streets, and that the grade 
of the curbs is flattened to eight per cent for the width of the intersecting 
sidewalks. Grades of less amount on. roadway or sidewalk are con- 
tinuous. The elevation of block-corners is found by adding together 
the curb elevations at the faces of the block-corners, and 2 per cent of 




the sum of the widths of the two sidewalks at the corner, and dividing 
the whole by two. This gives an elevation equal to the average eleva- 
tion of the curbs at the corners, plus an average rise of two and one- 
half per cent across the width of the sidewalk. 

Accommodation summits have to be introduced between street 
intersections first, in hilly localities, to avoid excessive excavation; 
and second, when the intersecting streets are level or nearly so, for the 
purpose of obtaining the fall necessary for surface drainage. 

The elevation and location of these summits may be calculated 
as follows : Let A be the elevation of the highest corner; B, the eleva- 
tion of the lowest corner; D, the distance from corner to corne?; 
and II, the rate of the accommodation grade. The elevation of the 
summit is equal to 

D-R + A + B. 


The distance from A or B is found by subtracting the elevation of 
either A or B from this quotient, and dividing the result by the rate 
of grade. Or the summits may be located mechanically by specially 
prepared scales. Prepare two scales divided to correspond to the rate 
of grade; that is, if the rate of grade be 1 foot per 100 feet, then one 
division of the scale should equal 100 feet on the map scale. These 
divisions may be subdivided into tenths. One scale should read from 
right to left, and one from left to right. 

To use the scales, place them on the map so that their figures 
correspond with the corner elevations; then, as the scales read in op- 

Curb Le</el_ 

* ~"~ oftom of Gutter **'** 

Fig. 49. 

posite directions, there is of course some point at which the opposite 
readings will be the same: this point is the location of the summits; 
and the figures read off the scale its elevation. If the difference in 
elevation of the corners is such as not to require an intermediate sum- 
mit for drainage, it will be apparent as soon as the scales are placed 
in position. 

When an accommodation summit is employed, it should be form- 
ed by joining the two straight grade lines by a vertical curve, as 




described in Part I. The curve should be used both in the crown of 
the street and in the curb and footpath. 

Where the grade is level between intersections, sufficient fall for 
surface drainage may be secured without the aid of accommodation 
summits, by arranging the grades as shown in Fig. 49. The curb is 
set level between the corners; a summit is formed in the gutter; and 
receiving basins are placed at each corner. 

Transverse Grade. In transverse grade the street should be 
level; that is, the curbs on opposite sides should be at the same level, 
and the street crown rise equally from each side to the center. But in 
hillside streets this condition cannot always be fulfilled, and opposite 

Fig. 52. 

sides of the street may differ as much as five feet; in such cases the 
engineer will have to use his discretion as to whether he shall adopt a 
straight slope inclining to the lo\ver side, thus draining the whole street 
by the lower gutter, or adopt the three-curb method and sod the slope 
of the higher side. 

In the improvement of old streets with the sides at different levels, 
much difficulty will be met, especially where shade trees have to be 
spared. In such cases, recognized methods have to be abandoned, and 
the engineer will have to adopt methods of overcoming the difficulties 
in accordance with the conditions and necessities of each particular 
case. Figs. 50, 51, and 52 illustrate several typical arrangements in 



the case of streets in which the opposite sides are at different levels. 
Transverse Contour or Crown. The reason for crowning a pave- 
ment i. e., making the center higher than the sides is to provide 
for the rapid drainage of the surface. The most suitable form for the 
crown is the parabolic curve, which may be started at the curb line, 
or at the edge of the gutter adjoining the carriage-way about one foot 

Fig. 53. 

from the curb. Fig. 53 shows this form, which is obtained by dividing 
the ordinate or width from the gutter to the center of the street into ten 
equal parts, and raising perpendiculars the length of which will be 
determined by multiplying the rise at the center by the respective 
number of each perpendicular in the diagram. The amounts thus 
obtained can be added to the rod readings; and the stakes, set at the 
proper distance across the street, with their tops at this level, will give 
the required curve. 

The amount of transverse rise, or the height of the center alx>ve 
the gutters, varies with the different paving materials, smooth pave- 
ments requiring the least, and rough ones and earth the greatest. The 
rise is generally stated in a proportion of the width of the carriage-way. 
The most suitable proportions are: 

Stone blocks, rise at center, J s width of carriage-way. 

Wood " " " T J n " " 

Brick " " " " ,,V " 

Asphalt" " " " t l ff 

Sub-Foundation Drainage of Streets. The sub-foundation 
drainage of streets cannot be effected by transverse drains, because of 
their liability to disturbance by the introduction of gas, water, and 
other pipes. 

Longitudinal drains must be depended upon entirely; they may 
be constructed of the same materials and in the same manner as road 
drains. The number of these longitudinal drains must tlepend upon 



the character of the soil. If the soil is moderately retentive, a single 
row of tiles or a .hollow invert placed under the sewer in the center of 
the street will generally be sufficient; or two rows of tiles may l)e em- 
ployed, one placed outside each curb line; if, on the other hand, the 
soil is exceedingly wet and the street very wide, four or more lines 
may be employed. These drains may be permitted to discharge into 
the sewers of the transverse streets. 

Surface Drainage. The removal of water falling on the street 
surface is provided for by collecting it in the gutters, from which it is 
discharged into the sewers or other channels by means of catch-basins 
placed at all street intersections and dips in the street grades. 

Gutters. The gutters must be of sufficient depth to retain all the 
water which reaches them and prevent its overflowing on the footpath. 
The depth should never be less than 6 inches, and very rarely need be 
more than 10 inches. 

Catch=basins are of various forms, usually circular or rectangular, 
built of brick masonry coated with a plaster of Portland cement. 
Whichever form is adopted, they should fulfil the following conditions: 

(1 ) The inlet and outlet should have sufficient capacity to receive 
and discharge all water reaching the basin. 

(2) The basins should have sufficient capacity below the outlet 
to retain all sand and road detritus, and prevent it being carried into 
the sewer. 

(3) They should be trapped so as to prevent the escape of sewer 
gas. (This requirement is frequently omitted, to the detriment of 
the health of the people.) 

(4) They should be constructed so that the pit can easily be 
cleaned out. 

(5) The inlet should be so constructed as not easily to be choked 
by leaves or debris. 

(6) They must offer the least possible obstruction to traffic. 

(7) The pipe connecting the basin to the sewer should be easily 
freed of any obstruction. 

* The bottom of the basins should be 6 or 8 feet below the street 
level ; and the water level in them should be from 3 to 4 feet lower than 
the street surface, as a protection against freezing. 

The capacity and number of basins will depend upon the area of 
surface which they drain 



In streets having level or light longitudinal grades, gullies may be 
formed along the line of the gutter at such intervals as may be found 

Catch-basins are usually placed at the curb line. In several cities, 
the basin is placed in the center of the street, and connects to 
inlets placed at the curb line. This reduces the cost of construction 
and cleaning, and removes from the sidewalk the dirty operations of 
cleaning the basins. 

Catch-basins and gully-pits require to be cleaned out at frequent 
intervals; otherwise the odor arising from the decomposing matter 
contained in them will be very offensive. No rule can be laid down 
for the intervals at which the cleaning should be done, but they must 
be cleaned often enough to prevent the matter in them from putrefying. 
There is no uniformity of practice observed by cities in this matter; in 
some, the cleaning is done but once a year; in others, after every rain- 
storm; in still others, at intervals of three or four months; w y hile in a 
few r cities the basins are cleaned out once a month. 


The stability, permanence, and maintenance of any pavement 
depend upon its foundation. If the foundation is weak, the surface 
will soon settle unequally, forming depressions and ruts. With a good 
foundation, the condition of the surface will depend upon the material 
employed for the pavement and upon the manner of laying it. 

The essentials necessary to the forming of a good foundation are : 

(1) The entire removal of all vegetable, perishable, and yielding 
matter. It is of no use to lay good material on a bad substratum. 

(2) The drainage of the subsoil wherever necessary. A per- 
manent foundation can be secured only by keeping the subsoil dry; 
for, where \vater is allowed to pass into and through it, its weak spots 
will be quickly discovered and settlement will take place. 

(3) The thorough compacting of the natural soil by .rolling with 
a roller of proper weight and shape until it forms a uniform and un- 
yielding surface. 

(4) The placing on the natural soil so compacted, a sufficient 
thickness of an impervious and incompressible material to cut off all 
communication between the soil and the bottom of the pavement. 

The character of the natural soil over \vhich the roadw T ay is to be 
built has an important bearing upon the kind of foundation and the 
manner of forming it; each class of soil will require its own special 



treatment. Whatever its character, it must be brought to a dry and 
tolerably hard -condition by draining and rolling. Sand and gravels 
which do not hold water, present no difficulty in securing a solid and 
secure foundation; clays and soils retentive of water are the most 
difficult. Clay should be excavated to a depth of at least 18 inches 
below the surface of the finished covering; and the space so excavated 
should be filled in with sand, furnace slag, ashes, coal dust, oyster 
shells, broken brick, or other materials which are not excessively absorb- 
ent of water. A clay soil or one retaining water may be cheaply and 
effectually improved by laying cross-drains with open joints at inter- 
vals of 50 or 100 feet. These drains should be not less than 18 inches 
below the surface, and the trenches filled with gravel. They should 
be 4 inches in internal diameter, and should empty into longitudinal 

Sand and planks, gravel, and broken stone have been successively 
used to form the foundation for pavements; but, although eminently 
useful materials, their application to this purpose has always been a 
failure. Being inherently weak and possessing no cohesion, the main 
reliance for both strength and wear must be placed upon the surface- 
covering. This covering usually (except in case of sheet asphalt) 
composed of small units, with joints between them varying from one- 
half an inch to one and a-half inches possesses no elements of cohe- 
sion; and under the blows and vibrations of traffic the independent 
units or blocks will settle and be jarred loose. On account of their 
porous nature, the subsoil quickly becomes saturated with urine and 
surface waters, which percolate through the joints; winter frosts up- 
heave them; and the surface of the street becomes blistered and broken 
up in dozens of places. 

Concrete. As a foundation for all classes of pavement (broken 
stone excepted), hydraulic-cement concrete is superior to any other. 
When properly constituted and laid, it becomes a solid, coherent mass 
capable of bearing great weight without crushing. If it fail at all, it 
must fail altogether. The concrete foundation is the most costly, but this 
is balanced by its permanence and by the saving in the cost of repairs to 
the pavement which it supports. It admits of access to subterranean 
pipes with less injury to the neighboring pavement than any other, for 
the concrete may be broken through at any point without unsettling 
the foundation for a considerable distance around it, as is the case with 



sand or other incoherent material; and when the concrete is replaced 
and set, the covering may be reset at its proper level, without the un- 
certain allowance for settlement which is necessary in other cases. 

Thickness of Concrete. The thickness of the concrete bed must 
be proportioned by the engineer; it should be sufficient to provide 
against breaking under transverse strain caused by the settlement of 
the subsoil. On a well-drained soil, six inches will be found sufficient; 
but in moist and clayey soils, twelve inches will not be excessive. On 
such soils a layer of sand or gravel, spread and compacted before pla- 
cing the concrete, will be found very beneficial. 

The proportions of the ingredients for concrete used for pavement 
foundations are usually: 

1 part Portland cement 
3 parts Sand 
7 parts Broken Stone. 

1 part Natural Hydraulic Cement 

2 parts Sand 

5 parts Broken Stone. 

The question is sometimes raised as to whether Natural or Port- 
land cement should be used. Natural cement is more extensively 
employed on account of its being cheaper in price than Portland. 
There is no advantage gained in using Portland cement. Concrete 
should not be laid when the temperature falls below 32 F. 

The concrete foundation, after completion, should be- allowed to 
remain several days before the pavement is placed upon it, in order 
that the mortar may become entirely set. During setting, the con- 
crete should be protected from the drying action of the sun and 
wind, and should be kept damp to prevent the formation of drying 


Stone blocks are commonly employed for pavements where traffic 
is heavy. The material of which the blocks are made should possess 
sufficient hardness to resist the abrasive action of traffic, and sufficient 
toughness to prevent them from being broken by the impact of loaded 
wheels. The hardest stones will not necessarily give the best re- 
sults in the pavement, since a very hard stone usually wears smooth 
and becomes slippery. The edges of the block chip off, and the 



upper face becomes rounded, thus making the pavement very 

The stone is sometimes tested to determine its strength, resistance 
to abrasion, etc. ; but, as the conditions of use are quite different from 
those under which it may be tested, such tests are seldom satisfactory. 
However, examination of a stone as to its structure, the closeness of its 
grain, its homogeneity, porosity, etc., may assist in forming an idea of 
its value for use in a pavement. A low degree of permeability usually 
indicates that the material will not be greatly affected by frost. 

Materials. Granite. Granite is more extensively employed for 
stone block paving than any other variety of stone; and because of this 
fact, the term "granite paving" is generally used as being synonymous 
with stone block paving. The granite employed should be of a tough, 
homogeneous nature. The hard, quartz granites are usually brittle, 
and do not wear well under the blows of horses' feet or the impact of 
vehicles; granite containing a high percentage of feldspar will be inju- 
riously affected by atmospheric changes; and granite in which mica 
predominates will wear rapidly on account of its laminated structure. 
Granite possesses the very important property of splitting in three 
planes at right angles to one another, so that paving blocks may readily 
be formed with nearly plane faces and square corners. This property 
is called the rift or cleavage. 

Sandstones of a close-grained, compact nature often give very 
satisfactory results under heavy traffic. They are less hard than 
granite, and wear more rapidly, but do not become smooth and slip- 
pery. Sandstones are generally known in the market by the name 
of the quarry or place where produced as "Medina," "Berea," 

Trap rock, while answering well the requirements as to durability 
and resistance to wear, is objectionable on account of its tendency to 
wear smooth and become slippery; it is also difficult to break into 
regular shapes. 

Limestone has not usually been successful in use for the construc- 
tion of block pavements, on account of its lack of durability against 
atmospheric influences. The action of frost commonly splits the 
blocks; and traffic shivers them, owing to the lamination being 



TABLE 12. 

Specific Gravity, Weight, Resistance to Crushing, and 
Absorption Power of Stones. 



per cu. ft. 

per sq. in. 





















Brick Paving 











Cobblestone Pavement. Cobblestones bedded in sand possess 
the merit of cheapness, and afford an excellent foothold for horses; 
but the roughness of such pavements requires the expenditure of a 
large amount of tractive energy to move a load over them. Aside from 
this, cobblestones are entirely wanting in the essential requisites of a 
good pavement. The stones being of irregular size, it is almost impos- 
sible to form a bond or to hold them in place. Under the action of the 
h affic and frost, the roadway soon becomes a mass of loose stones. 
Moreover, cobblestone pavements are difficult to keep clean, and very 
unpleasant to travel over. 

Belgian Block Pavement. Cobblestones were displaced by pave- 
ments formed of small cubical blocks of stone. This type of 
pavement was first laid in Brussels, thence imported to Paris, and from 
there taken to the United States, where it has been widely known as 
the "Belgian block" pavement. It has been largely used in New York 
City, Brooklyn, and neighboring towns, the material being trap-rock 
obtained from the Palisades on the Hudson River. 

The stones, being of regular shape, remain in place better than 
cobblestones; but the cubical form (usually five inches in each dimen- 
sion) is a mistake. The foothold is bad; the stones wear round; and 
the number of joints is so great that ruts and hollows are quickly 
formed. This pavement offers less resistance to traction than cobble- 
stones, but it is almost equally rough and noisy. 

Granite Block Pavement. The Belgian block has been gradually 



displaced by the introduction of rectangular blocks of granite. Blocks 
of comparatively large dimensions were at first employed. They were 
from 6 to 3 inches in width on the surface, from 10 to 20 inches in 
length, with a depth of 9 inches. They were merely placed in rows 
on the subsoil, perfunctorily rammed, the joints filled with sand, and 
the street thrown open to traffic. The unequal settlement of the 
blocks, the insufficiency of the foothold, and the difficulty of cleansing 
the street, led to the gradual development of the latest type of stone- 
block pavement, which consists of narrow, rectangular .blocks of 
granite, properly proportioned, laid on an unyielding and impervious 
foundation, with the joints between the blocks filled with an imper- 
meable cement. 

Experience has proved beyond doubt that this latter type of 
pavement is the most enduring ai d economical for roadways subjected 
to heavy and constant traffic. Its advantages are many, while its 
defects are few. 


(1) Adaptability to all grades. 

(2) Suits all classes of traffic. 

(3) Exceedingly durable. 

(4) Foothold, fair. 

(5) Requires but little repair. 

(6) Yields but little dust or mud. 

(7) Facility for cleansing, fair. 

(1) Under certain conditions of the atmosphere, the surface of 
the pavement becomes greasy and slippery. 

(2) The incessant din and clatter occasioned by the movement 
of traffic is an intolerable nuisance; it is claimed by many physicians 
that the noise injuriously affects the nerves and health of persons who 
are obliged to live or do business in the vicinity of streets so paved. 

(3) Horses constantly employed upon it soon suffer from the 
continual jarring produced in their legs and hoofs, and quickly wear 

(4) The discomfort of persons riding over the pavement is very 
great, because of the continual jolting to which they are subjected. 

(5) If stones of an unsuitable quality are used for example, 



those that polish the surface quickly becomes slippery and exceed- 
ingly unsafe for travel. 

Size and Shape of Blocks. The proper size of blocks for paving 
purposes has been a subject of much discussion, and a great variety of 
forms and dimensions are to be found in all cities. 

For stability, a certain proportion must exist between the depth, 
the length, and the breadth. The depth must be such that when the 
wheel of a loaded vehicle passes over one edge of the upper surface 
of a block, the block will not tend to tip up. The resultant direction 
of the pressure of the load and adjoining blocks should always tend 
to depress the whole block vertically ; where this does not happen, the 
maintenance of a uniform surface is impossible. To fulfil this require- 
ment, it is not necessary to make the block more than six inches deep. 

Width of Blocks. The maximum width of blocks is controlled 
by the size of horses' hoofs. To afford good foothold to horses draw- 
ing heavy loads, it is necessary that the width of each block, measured 
along the street, shall be the least possible consistent with stability. 

Gutter formed of 3 rows of* 
ti/ocks, Set long/tuciirialy 

Fig. 54. 

If the width be great, a horse drawing a heavy load, attempting to find 
a joint, slips back, and requires an exceptionally wide joint to pull him 
up. It is therefore desirable that the width of a block shall not exceed 
3 inches ; or that four blocks, taken at random and placed side by side, 
shall not measure more than 14 inches. 

Length of Blocks, The length, measured across the street, 
must be sufficient to break joints properly, for two or more joints in 
line lead to the formation of grooves. For this purpose the length 
of the block should be not less than 9 inches nor more than 12 inches. 

Form of Blocks. The blocks should be well squared, and must 
not taper in any direction ; sides and ends should be free from irregular 
projections. Blocks that taper from the surface downwards (wedge- 
shaped) should not be permitted in the work; but if any are allowed, 
they should be set with the widest side down. 




Alanner of Laying Blocks. The blocks should be laid in parallel 
courses, with their longest side at right angles to the axis of the street, 
and the longitudinal joints broken by a lap of at least two inches (see 
Figs. 54 and 55). The reason for this is to prevent the formation of 
longitudinal ruts, which would happen if the blocks were laid length- 
wise. Laying blocks obliquely and "herring-bone" fashion has been 

tried in several cities, with the idea that the wear and formation of ruts 
would be reduced by having the vehicle cross the blocks diagonally. 
The method has failed to give satisfactory results; the wear was ir- 
regular and the foothold defective; the difficulty of construction was 
increased by reason of labor required to form the triangular joints; and 
the method was wasteful of material. 

Fig. 56. 

The gutters should be formed by three or more courses of block, 
laid with their length parallel to the curb- 

At junctions or intersections of streets, the blocks should be laid 
diagonally from the center, as shown in Fig. 56. The reasons for 



this are: (1) To prevent the traffic crossing the intersection from 
following the longitudinal joints and thus forming depressions and 
ruts; (2) Laid in this manner, the blocks afford a more secure foot- 
hold for horses turning the corners. The ends of the diagonal blocks 
where they abut against the straight blocks, must be cut to the re- 
quired bevel. 

The blocks forming each course must be of the same depth, and 
no deviation greater than one-quarter of an inch should be permitted. 
The blocks should be assorted as they are delivered, and only those 
corresponding in depth and width should be used in the same course. 
The better method would be to gauge the blocks at the quarry. 
This would lessen the cost considerably ; it would also avoid the in- 
convenience to the public due to the stopping of travel because of the 
rejection of defective material on the ground. This method would 
undoubtedly be preferable to the contractor, who would be saved the 
expense of handling unsatisfactory material ; and it would also leave 
the inspectors free to pay more attention to the manner in which the 
work of paving is performed. 

The accurate gauging of the blocks is a matter of much impor- 
tance. If good work is to be executed, the blocks, when laid, must be 
in parallel and even courses; and if the blocks be not accurately gauged 
to one uniform size, the result will be a badly paved street, with the 
courses running unevenly. The cost of assorting blocks into lots of 
uniform wjdth, after delivery on the street, is far in excess of any ad- 
ditional price which would have to be paid for accurate gauging at the 

Foundation. The foundation of the blocks must be solid and 
unyielding. A bed of hydraulic-cement concrete is the most suitable, 
the thickness of which must be regulated according to the traffic; the 
thickness, however, should not be less than 4 inches, and need not be 
more than 9 inches. A thickness of 6 inches will sustain traffic of 600 
tons per foot of width. 

Cushion Coat. Between the surface of the concrete and the base 
of the blocks, there must be placed a cushion-coat formed of an incom- 
pressible but mobile material, the particles of which will readily adjust 
themselves to the irregularities of the bases of the blocks and transfer 
the pressure of the traffic uniformly to the concrete below. A layer 
pf dry, clean sand 1 to 2 inches thick forms an excellent cushion-coat- 



Its particles must be of such fineness as to pass through a No. 8 screen; 
if coarse and containing pebbles, they will not adapt themselves to the 
irregularities of the bases of the blocks; hence the blocks will be sup- 
ported at only a few points, and unequal settlement will take place when 
the pavement is subjected to the action of traffic. The sand must also 
be perfectly free from moisture, and artificial heat must be used to dry 
it if necessary. This requirement is an absolute necessity. There 
should be no moisture below the blocks when laid ; nor should water 
be allowed to penetrate below the blocks; if such happens, the effect of 
frost will be to upheave the pavement and crack the concrete. 

Where the best is desired without regard to cost, a layer half an 
inch thick of asphaltic cement may be substituted for the sand, with 
superior and very satisfactory results. 

Laying Blocks. The blocks should be laid stone to stone, so that 
the joint may be of the least possible width; wide joints cause increased 
wear and noise, and do not increase the foothold. The courses should 
be commenced on each side and worked toward the middle; and the 
last stone should fit tightly. 

Ramming. After the blocks have been set, they should be well 
rammed down ; and the stones which sink below the general level 
should be taken up and replaced with a deeper stone or brought to 
level by increasing the sand bedding. 

The practice of workmen is invariably to use the rammer so as to 
secure a fair surface. This is not the result intended to be secured, 
but to bring each block to an unyielding bearing. The result of such 
a surfacing process is to produce an unsightly and uneven roadway 
when the pressure of traffic is brought upon it. The rammer used 
should weigh not less than 50 pounds and have a diameter of not less 
than 3 inches. 

Joint Filling. All stone block pavements depend for their water- 
proof qualities upon the character of the joint filling. Joints filled 
with sand and gravel are of course pervious. A grout of lime or cement 
mortar does not make a permanently waterproof joint; it becomes 
disintegrated under the vibration of traffic. An impervious joint can 
be made only by employing a filling made from bituminous or asphaltic 
material; this renders the pavement more impervious to moisture, 
makes it less noisy, and adds considerably to its strength. 

Bituminous Cement for Joint Filling. The bituminous materials 



employed are: (1) The tar produced in the manufacture of gas, 
which, when redistilled, is called distillate, and is numbered 1, 2, 3, 4, 
etc., according to its density; this, material under the name of paving- 
pitch is extensively used, both alone and in combination with other 
bituminous substances; (2) Combinations of gas tar or coal tar with 
refined asphaltum; (3) Mixtures of refined asphaltum, creosote, and 
coal tar. 

The formula for the bituminous joint rilling used in New York 
City, is: 

Refined Trinidad asphaltum 20 parts. 

No. 4 coal-tar distillate 100 parts. 

Residuum of petroleum 3 parts. 

In Washington, D. C., coal tar distillate No. 6 is used alone. 

In Europe a bituminous cement much used is composed of coal- 
tar, asphaltum, gas tar, and creosote oil, in the proportion of 100 
pounds of asphaltum to 4 gallons of tar and 1 gallon of creosote. 
These proportions are varied somewhat, according to the quality of the 
asphaltum employed. The mixture is melted, and is boiled from 
one to two hours in a suitable boiler, being then poured into the joints 
in a boiling state. This mixture is impervious to moisture, and pos- 
sesses a degree of elasticity sufficient to prevent it from cracking. 

The mode of applying the bituminous cement is as follows : After 
the blocks are rammed, the joints are filled to a depth of about two 
inches with clean gravel heated to a temperature of about 250 F. ; 
then the hot cement is poured in until it forms a layer of about one inch 
on top of the gravel; then more gravel is filled in to a depth of about 
two inches; then cement is poured in until it appears on top of the 
gravel, more gravel being next added until it reaches to within half an 
inch of the top of the blocks; this remaining half-inch is filled with 
cement, and then fine gravel or sand is sprinkled over the joints. 

In some cases the joints are first filled with heated gravel; the 
cement is poured in until the sand beneath and the gravel between 
the blocks will absorb no more, and the joints are filled flush with the 
top of the pavement. This method is open to objection; for, if the 
gravel is not sufficiently hot, the cement will be chilled and will not flow 
to the bottom of the joint, but, instead, will form a thin layer near the 
surface, which under the action of frost and the vibration of traffic, 
will be quickly cracked and broken up; the gravel will settle, and the 


View near Eide, on Hardanger Fjord, Norway. 


blocks will be jarred loose, the surface of the pavement becoming a 
series of ridges and hollows. 

The quantity of cement required per square yard of pavement will 
vary according to the shape of the blocks, the width of the joints, and 
the depth of the sand bed. With well-shaped blocks, close joints, and 
a half-inch sand bed, the quantity will vary from 3V to 5 gallons; with 
ill-shaped blocks, wide joints, and a heavy sand bed, 10 to 12 gallons 

Fig. 57. 

would not be an excessive amount to use to secure the result obtained 
by employing well-shaped blocks and close joints. 

Stone Pavement on Steep Grades. Stone blocks may be em- 
ployed on all practicable grades; but on grades exceeding 10 per cent, 
cobblestones afford a better foothold than blocks. The cobblestones 
should be of uniform length, the length being at least twice the breadth 
say stones 6 inches long and 2^ to 3 inches in diameter. These 
should be set on a concrete foundation, laid stone to stone, and the 



Fig. 58. 

interstices filled with cement grout or bituminous cement; or a bitu- 
minous concrete foundation may be employed and the interstices be- 
tween the stones filled with asphaltic paving cement. Should stone 
blocks be preferred, they must be laid, when the grade exceeds 5 per 
cent, with a serrated surface, by either of the methods shown in Figs. 
57 and 58. The method shown in Fig. 57 consists in slightly tilting 
the blocks on their bed so as to form a series of ledges or steps, against 
which the horses' feet being planted, a secure foothold is obtained. 
The method shown in Fig. 58 consists in placing between the rows of 



stones a course of slate, or strips of creosoted wood, rather less than 
one inch in thickness and about an inch less in depth than the blocks; 
or the blocks may be spaced about one inch apart, and the joints filled 
with a grout composed of gravel and cement. The pebbles of the 
gravel should vary in size between one-quarter and three-quarters of 
an inch. 


Characteristics of Brick Suitable for Paving. These are: 

(1) Not to be acted upon by acids. 

(2) Not to absorb more than 1-600 of its weight of water in 
48 hours. 

(3) Not susceptible to polish. 

(4) Rough to the touch, resembling fine sandpaper. 

(5) To give a clear, ringing sound when struck together. 

(6) When broken, to show a compact, uniform, close-grained, 
structure, free from air-holes and pebbles. 

(7) Not to scale, spall, or chip when quickly struck on the edges. 

(8) Hard but not brittle. 

Tests of Paving Brick. To ascertain the quality of paving brick, 
they are generally subjected to four tests, namely: (1) Abrasion by 
impact (commonly called the "Rattler" test); (2) absorption; (3) 
transverse or cross-breaking; (4) crushing. With a view to securing 
uniformity in the methods of making the above tests, the National 
Brick Manufacturers' Association has adopted and recommends the 

Rattler Test 

1. Dimensions of the Machine. The standard machine shall 
be 28 inches in diameter and 20 inches in length, measured inside the 

Other machines may be used, varying in diameter between 26 and 
30 inches, and in length between 18 and 24 inches; but if this is done, 
a record of it must be attached to the official report. I^ong rattlers 
may be cut up into sections of suitable length by the insertion of an 
iron diaphragm at the proper point. 

2 Construction of the Machine. The barrel shall be supported 
on trunnions at either end; in no case shall a shaft pass through the 
rattling chamber. The cross-section of the barrel shall be a regular 
polygon having 14 sides. The heads shall be composed of gray cast- 


iron, not chilled or case-hardened. The staves shall preferably be 
composed of steel plates, since cast-iron peans and ultimately breaks 
under the wearing action on the inside. There shall be a space of one- 
fourth of an inch between the staves, for the escape of dust and small 
pieces of waste. Other machines may be used, having from twelve to 
sixteen staves; but if this is done, a record of it must be attached to 
the official report of the test. 

3. Composition of the Charge. All tests must be executed on 
charges containing but one make of brick or block at a time. The 
charge shall consist of 9 paving blocks or 12 paving bricks, together 
with 300 pounds of shot made of ordinary machinery cast-iron. This 
shot shall be of two sizes, as described below; and the shot charge shall 
be composed one-fourth (75 pounds) of the larger size, and three- 
fourths (225 pounds) of the smaller size. 

4. Size of the Shot. The larger size shall weigh about 1\ pounds 
and be about 2i inches square and 4J- inches long, with slightly round- 
ed edges. The smaller size shall be cubes of \\ inches on a side, with 
rounded edges. The individual shot shall be replaced by new .ones 
when they have lost one-tenth of their original weight. 

5. Revolutions of the Charge. The number of revolutions of a 
standard test shall be 1,800; and the speed of rotation shall not fall 
below 28 nor exceed 30 revolutions per minute. The belt-power shall 
be sufficient to rotate the rattler at the same speed, whether charged 
or empty. 

(>. Condition of the Charge. The bricks composing a charge 
shall be thoroughly dried before making the test. 

7. Calculation of the Results. The loss shall be calculated in 
per cents of the weight of the dry brick composing the charge; and no 
result shall be considered as official unless it is the average of two 
distinct and complete tests made on separate charges of brick. 
Absorption Test 

1. The number of bricks for a standard test shall be five. 

2. The test must be conducted on rattled brick. If none such 
are available, the whole brick must be broken in halves before treatment. 

3. Dry the bricks for 48 hours at a temperature ranging from 
230 to 250 F. before weighing for the official dry weight. 

4. Soak for 48 hours completely immersed in pure water. 



5. After soaking, and before weighing, the bricks must be wiped 
dry from surplus water. 

6. The difference in the weight must be determined on scales 
sensitive to one gram. 

7. The increase in weight due to water absorbed shall be ca.- 
culated in per cents of the initial dry 7 weight. 

Cross=Breaking Test 

1. Support the brick on edge, or as laid in the pavement, on 
hardened steel knife-edges, rounded longitudinally to a radius of 
twelve inches and transversely to a radius of one-eighth inch, and 
bolted in position so as to secure a span of six inches. 

2. Apply the load to the middle of the top face through a hard- 
ened steel knife-edge, straight longitudinally and rounded transversely 
to a radius of one-sixteenth inch. 

3. Apply the load at a Uniform rate of increase till fracture 

4. Compute the modulus of rupture by the formula 

' 2bd?' 

in which / = modulus of rupture, in pounds per square inch ; 
w = total breaking load, in pounds; 
/ = length of span, in inches = 6 ; 
b = breadth of brick, in inches- 
d = depth of brick, in inches. 

5. Samples for test must be free from all visible irregularities of 
surface or deformities of shape, and their upper and lower faces must 
be practically parallel. 

6. Not less than ten brick shall be broken, and the average of all 
shall be taken for a standard test. 

Crushing Test 

1. The crushing test should be made on half-bricks, loaded 
edgewise, or as they are laid in the street. If the machine used is 
unable to crush a full half-brick, the area may be reduced by chipping 
off, keeping the form of the piece to be tested as nearly prismatic as 
possible. A machine of at least 100,000 pounds' capacity should be 
used; and the specimen should not be reduced below four square 
inches of area in cross-section at right angles to direction of load. 

2. The upper and lower surfaces should preferably be ground to 



true and parallel planes. If this is not done, they should be bedded, 
while in the testing machine, in plaster of Paris, which should be 
allowed to harden ten minutes under weight of the crushing planes 
only, before the load is applied. 

3. The load should be applied at a uniform rate of increase to 
the point of rupture. 

4. Not less than an average obtained from five tests on five 
different bricks shall constitute a standard test. 

Properties of Paving Bricks. Paving bricks range in weight 
from 5?, to 7\ pounds; in specific gravity, from 1.91 to 2.70; in resist- 
ance to crushing, from 7,000 to 18,000 pounds per square inch; in 
resistance to cross-breaking, R = 1,400 to 2,000 pounds; in absorption, 
from 0.15 to 3 per cent in 24 hours. The dimensions vary according to 
locality and the requirements of the specifications. The "standard" 
bricks are 21 X 4 X 8 inches, requiring 58 bricks to the square yard, 
and weighing 7 pounds each; "repressed", 2 X 4 X8 inches, requir- 
ing 61 to the square yard, and weighing 6V pounds each; "Metropoli- 
tan", 3X4X9 inches, requiring 45 to the square yard, and weighing 
9 2 pounds each. 

Advantages of Brick Pavements. These may be stated as follows: 

(1) Ease of traction. 

(2) Good foothold for horses. 

(3) Not disagreeably noisy. 

(4) Yields but little dust and mud. 

(5) Adapted to all grades. 

(6) Easily repaired. 

(7) Easily cleaned. 

(8) But slightly absorbent. 

(9) Pleasing, to the eye. 

(10) Expeditiously laid. 

1) Durable under moderate traffic. 

Defects of Brick Pavements. The principal defects of brick 
pavements arise from lack of uniformity in the quality of the bricks, 
and from the liability of incorporating in the pavement bricks of too 
soft or porous a structure, which crumbles under the action of traffic 
or frost. 

Foundation. A brick pavement should have a firm foundation. 
As the surface is made up of small, independent blocks, each one must 



be adequately supported, or the load coming upon it may force it 
downwards and cause unevenness, a condition which conduces to the 
rapid destruction of the pavement. Several forms of foundation have 
been used such as gravel, plank, sand, broken stone, and concrete. 
The last mentioned is doubtless the best. 

Sand Cushion. The sand cushion is a layer of sand placed on 
top of the concrete to form a bed for the brick. Practice regarding 
the depth of this layer of sand varies considerably. In some cases it 
is only half an inch deep, varying from this up to three inches. The 
sand cushion is very desirable, as it not only forms a perfectly true and 
even surface upon which to place brick, but also makes the pavement 
less hard and rigid than would be the case were the brick laid directly 
on the concrete. 

The sand is spread evenly, sprinkled with water, smoothed, and 
brought to the proper contour by screeds or wooden templets, properly 
trussed, mounted on wheels or shoes which bear upon the upper sur- 
face of the curb. Moving the templet forward levels and forms the 
sand to a uniform surface and proper shape. 

The sand used for the cushion-coat should be clean and free from 
loam, moderately coarse, and free from pebbles exceeding one-quarter 
inch in size. 

Manner of Laying. The bricks should be laid oh edge, as closely 
and compactly as possible, in straight courses across the street, with 
the length of the bricks at right angles to the axis of the street. Joints 
should be broken by at least 3 inches. None but whole bricks should 
be used, except in starting a course or making a closure. To provide 
for the expansion of the pavement, both longitudinal and transverse 
expansion-joints are used, the former being made by placing a board 
templet seven-eighths of an inch thick against the curb and abutting 
the brick thereto. The transverse joints are formed at intervals 
varying between 25 and 50 feet, by placing a templet or building-lath 
three-eighths of an inch thick between two or three rows of brick. 
After the bricks are rammed and ready for grouting, these templets are 
removed, and the spaces so left are filled with coal-tar pitch or asphal- 
tic paving cement. The amount of pitch or cement required will vary 
between one and one and a-half pounds per square yard of pavement, 
depending upon the width of the joints. After 25 or 30 feet of the 
pavement is laid, every part of it should be rammed with a rammer 


weighing not less than 50 pounds; and the bricks which sink below the 
general level should be removed, sufficient sand being added to raise 
the brick to the required level. After all objectionable brick have 
been removed, the surface should be swept clean, then rolled with a 
steam roller weighing from 3 to 6 tons. The object of rolling is to 
bring the bricks to an unyielding bearing with a plane surface; if this 
is not done, the pavement will be rough and noisy and will lack dura- 

Fig. 59. 

bility. The rolling should be first executed longitudinally, beginning 
at the crown and working toward the gutter, taking care that each 
return trip of the roller covers exactly the same area as the preceding 
trip, so that the second passage may neutralize any careening of the 
brick due to the first passage. 

The manner of laying brick at street intersections is shown in 
Figi 59. 



Joint Filling. The character of the material used in filling the 
joints between the brick lias considerable influence on the success and 
durability of the pavement. Various materials have been used such 
as sand, coal-tar pitch, asphalt, mixtures of coal-tar and asphalt, and 
Portland cement, besides various patented fillers ; as "Murphy's 
grout", which is made from ground slag and cement. Each material 
has its advocates, and there is much difference of opinion as to which 
gives the best results. 

The best results seem to be obtained by using a high grade of 
Portland cement containing the smallest amount of lime in its composi- 
tion, the presence of the lime increasing the tendency of the filler to 
swell through absorption of moisture, causing the pavement to rise 
or to be lifted away from its foundation, and thus producing the roaring 
or rumbling noise so frequently complained of. 

The Portland cement grout, when uniformly mixed and carefully 
placed, resists the impact of traffic and wears well with brick. When a 
failure occurs/repairs can be made quickly; and, if made early, the 
pavement 'will be restored to a good condition. If, however, repairs 
are neglected, the brick soon loosens and the pavement fails. 

The office of a filler is to prevent water from reaching the founda- 
tion, and to protect the edges of the brick from spalling under traffic. 
In order to meet both of these requirements, every joint must be filled 
to the top, and must remain so, wearing down with the brick. Sand 
does not meet these requirements. Although at first making a good 
filler, being inexpensive and reducing the liability of the pavement to 
be noisy, it soon washes out, leaving the edges of the brick unprotected 
and consequently liable to be chipped. Coal-tar and the mixtures of 
coal-tar and asphalt have an advantage in rendering a pavement less 
noisy and in cementing together any breaks that may occur through 
upheavals from frost or other causes ; but, unless made very hard, they 
have the disadvantage of becoming soft in hot weather and flowing 
to the gutters and low places in the pavement, there forming a black 
and unsightly scale and leaving the high parts unprotected. The 
joints, thus deprived of their filling, become receptacles for water, mud, 
and ice in turn; and the edges of the brick are quickly broken down. 
Some of these mixtures become so brittle in winter that they crack 
and fly out of the joints under the action of traffic. 

The Portland cement filler is prepared by mixing two parts of 



cement and one part of fine sand with sufficient water to make a thin 
grout. The most convenient arrangement for preparing and dis- 
tributing the grout is a water-tight wooden box carried on four wooden 
wheels about 12 inches in diameter. The box may be about 4 feet 
wide, 7 feet long, and 12 inches deep, furnished with a gate about 8 
inches wide, in the rear end. The box should be mounted on the 
wheels with an inclination, so that the rear end is about. 4 inches lower 
than the front end. 

The operation of placing the filler is as follows : The cement and 
sand are placed in the box, and sufficient water is added to make a 
thin grout. The box is located about 12 feet from the gutter, the end 
gate opened, and about 2. cubic feet of the grout allowed to flow out 
and run over the top of the brick (care being taken to stir the grout 
while it is being discharged). If the brick are very "dry, the entire 
surface of the pavement should be thoroughly wet with a hose before 
applying the grout; if not, absorption of the water from the grout by 
the bricks will prevent adhesion between the bricks and the cement 
grout. The grout is swept into the joints by ordinary bass brooms. 
After about 100 feet in length of the pavement has been covered the 
box is returned to the starting-point, and the operation is repeated 
with a grout somewhat thicker than the first. If this second applica- 
tion is not sufficient to fill the joints, the operation is repeated as often 
as may be necessary to fill them. If the grout has been made too thin, 
or the grade of the street is so great that the grout will not remain long 
enough in place to set, dry cement may be sprinkled over the joints and 
swept in. After the joints are completely filled and inspected, allowing 
three or four hours to intervene, the completed pavement should be 
covered with sand to a depth of about half an inch, and the roadway 
barricaded, and no traffic allowed on it for at least ten days. 

The object of covering the pavement with sand is to prevent the 
grout from drying or settling too rapidly; hence, in dry and windy 
weather, it should be sprinkled from time to time. If coarse sand is 
employed in the grout, it will separate from the cement during the 
operation of filling the joints, with the result that many joints will be 
filial with sand and very little cement, while others will be filled with 
cement and little or no sand; thus there will be many spots in the pave- 
ment in which no bond is foimed between the bricks, and under the 
action of traffic these portions will quickly become defective. 



The coal-tar filler is best applied by pouring the material from 
buckets, and brooming it into the joints with wire brooms. In order to 
fill the joints effectually, it must be used only when very hot. To 
secure this condition, a heating tank on wheels is necessary. It should 
have a capacity of at least five barrels, and be kept at a uniform tem- 
perature all day. One man is necessary to feed the fire and draw the 
material into the buckets; another, to carry the buckets from the heat- 
ing-tank to a third, who pours the material over the street. The latter 
starts to pour in the center of the street, working backward toward the 
curb, and pouring a strip about two feet in width. A fourth man, 
with a wire broom, follows immediately after him, sweeping the sur- 
plus material toward the pourer and in the direction of the curb. This 
method leaves the entire surface of the pavement covered with a thin 
coating of pitch, which should immediately be covered with a light 
coating of sand; the sand becomes imbedded in the pitch. Under the 
action of traffic, this thin coating is quickly worn away, leaving the 
surface of the bricks clean and smooth. 

Tools Employed in Construction of Block Pavements. The 
principal tools required in constructing block pavements comprise 
hammers and rammers of varying sizes and shapes, depending on the 
material and size of the blocks to be laid; also crowbars, sand screens, 
and rattan and wire brooms. Cobblestones, square blocks, and brick 
require different types of both hammers and rammers for adjusting 
them to place and forcing them to their seat. A cobblestone rammer, 
for example, is usually made of wood (generally locust) in the shape of 
a long truncated cone, banded with iron at top and bottom, weighing 
about 40 pounds, and having two handles, at the top and another 
'on one side. A Belgian block rammer is slightly heavier, consisting 
of an upper part of wood set in a steel base; while a rammer for granite 
blocks is still heavier, comprising an iron base with cast-steel face, into 
which is set a locust plug with hickory handles. For laying brick, a 
wooden rammer shod with cast iron or steel and weighing about 27 
pounds is used. A light rammer of about 20 pounds' weight, consist' 
ing of a metallic base attached to a long, slim wooden handle, is used 
for miscellaneous work, such as tamping in trenches, next to curbs, etc. 

Concrete=Mixing Machine. Where large quantities of concrete 
are required, as in the foundations of improved pavements, concrete 
can be prepared more expeditiously and economically by the use of 




mechanical mixers, and the ingredients will he more thoroughly mixed, 
than by hand. Thorough incorporation of the ingredients is an essen- 
tial element in the quality of a concrete. When mixed by hand, how- 
ever, the incorporation is rarely complete, because it depends upon the 
proper manipulation of the hoe and shovel. The manipulation, 
although extremely simple, is rarely performed by the ordinary laborer 
unless he is constantly watched by the overseer. 

Several varieties of concrete-mixing machines are in the market. 
A convenient portable type is illustrated in Fig. 60. The capacity of 

Fig. 60. Concrete Mixing Machine. 

the mixers ranges from five to twenty cubic yards per hour, depending 
upon size, regularity with which the materials are supplied, speed, etc. 

Gravel Heaters. Fig. 61 illustrates a device commonly employed 
for heating the gravel used for joint filling in stone-block pavements. 
These heaters are made in various sizes, a common size being 9 feet 
long, 5 feet wide, and 3 feet 9 inches high. 

Melting Furnaces, for heating the pitch or tar for joint filling, are 
illustrated by Fig. 62. Various sizes are on the market. 

Wood pavements are formed of either rectangular or cylindrical 
blocks of wood. The rectangular blocks are generally 3 inches wide, 



9 inches long, and G inches deep; the round blocks are commonly 6 
inches in diameter and 6 inches long. 

The kinds of wood most commonly used are cedar, cypress, juni- 
per, yellow pine, and mesquite; and recently jarrah from Australia, 
and pyinyado from India, have been used. 

The wood is used in its natural condition, or impregnated with 
creosote or other chemical preservative. 

The blocks of wood are laid either .on the natural soil, on a bed 
of sand and gravel, on a layer of broken stone, on a layer of concrete, 

Melting Furnace. 

or, sometimes, on a double layer of plank. The joints are filled either 
with sand, paving-pitch, or Portland-cement grout. 

Advantages. The advantages of wood pavement may be stated 
as follows: 

(1) It affords good foothold for horses. 

(2) It offers less resistance to traction than stone, and slightly 
more than asphalt. 

(3) It suits all classes of traffic. 

(4) It may be used on grades up to five per cent. 

(5) It is moderately durable. 

(6) It yields no mud when laid upon an impervious foundation. 

(7) It yields but little dust. 



(8) It is moderate in first cost. 

(9) It is not disagreeably noisy. 

Defects. The principal objections to wood pavement are: 

(1) It is difficult to cleanse. 

(2) tnder certain conditions of the atmosphere it becomes 
greasy and very unsafe for horses. 

(3) It is not easy to open for the purpose of gaining access to 
underground pipes, it being necessary to remove rather a large surface 
for this purpose, which has to be left a little time after being repaired 
before traffic is again allowed upon it. 

(4) It is absorbent of moisture. 

(5) It is claimed by many that wood pavements are unhealthy. 
Quality of Wood. The question as to which of the various kinds 

of wood available is the most durable and economical, has not 
been satisfactorily determined. Many varieties have been tried. In 
England, preference is given to Baltic fir, yellow pine, and Swedish 
yellow deal. In the United States the variety most used (on account 
of its abundance and cheapness) is cedar; but yellow pine, tamarack, 
and mesquite have also been used to a limited extent, and cypress and 
juniper are being largely used in some of the Southern States. 

Hardwoods, such. as oak, etc., do not make the best pavements, as 
such woods become slippery. The softer, close-grained woods, such 
as cedar and pine, wear better and give good foothold. 

The wood employed should be sound and seasoned, free from sap, 
shakes, and knots. Defective blocks laid in the pavement will quickly 
cause holes in the surface, and the adjoining blocks will suffer tinder 
wear, the whole surface becoming bumpy. 

Chemical Treatment of Wood. The great enemy of all wood 
pavements is decay, induced by the action of the air and water. Wood 
is porous, absorbs moisture, and thus hastens its own destruction. 
Many processes have been invented to overcome this defect. The 
most popular processes at present are creosoiing and modifications of 
the same, known as the "creo-resinate" and "kreodine" processes. 
These consist of creosote mixed with various chemicals which are 
supposed to add to the preserving qualities of the creosote. 

Creosoting. This process consists in impregnating the wood with 
the oil of tar, called creosote, from which the ammonia has been ex- 
pelled, the effect being to coagulate the albumen and thereby prevent 



its decomposition, also to fill the pores of the wood with a bituminous 
substance which excludes both air and moisture, and which is noxious 
to the lower forms of animal and vegetable life. In adopting this pro- 
cess, all moisture should be dried out of the pores of the timber. The 
softer woods, while warm from the drying-house, may be immersed at 
once in an open tank containing hot creosote oil, when they will absorb 
about 8 or 9 pounds per cubic foot. For hardwoods, and woods which 
are required to absorb more than 8 or 9 pounds of creosote per cubic 
foot, the timber should be placed in an iron cylinder with closed ends, 
and the creosote, which should be heated to a temperature of about 
120 F., forced in with a pressure of 170 pounds to the square inch. 
The heat must be kept up until the process is complete, to prevent the 
creosote from crystallizing in the pores of the wood. By this means 
the softer woods will easily absorb from 10 to 12 pounds of oil per cubic 

The most effective method, however, is to exhaust the air from tht 
cylinder after the timber is inserted ; then to allow the oil to flow in ; and 
when the cylinder is full, to use a force pump with a pressure of 150 to 
200 pounds per square inch, until the wood has absorbed the requisite 
quantity of oil, as indicated by a gauge, which should be fitted to the 
reservoir tank. 

The oil is usually heated by coils of pipe placed in the reservoir, 
through which a current of steam is passed. 

The quantity of creosote oil recommended to be forced into the 
wood is from 8 to 12 pounds per cubic foot. Into oak and other hard 
woods it is difficult to force, even with the greatest pressure, more than 
2 or 3 pounds of oil. 

The advantages of this process are : The chemical constituents of 
the oil preserve the fibers of the wood by coagulating the albumen of 
the sap; the fatty matters act mechanically by filling the pores and thus 
exclude water; while the carbolic acid contained in the oil is a powerful 
disinfectant. . 

The life of the wood is extended by any of the above processes, by 
preserving it from decay; but such processes have little or no effect on 
the wear of the blocks under traffic. 

The process of dipping the blocks in coal tar or creosote oil is 
injurious. Besides affording a cover for the use of defective or sappy 
wood, it hastens decay, especially of green wood; it closes up the ex- 




terior of the cells of the wood so that moisture cannot escape, thus 
causing fermentation to take place in the interior of the block, which 
quickly destroys the strength of the fibers and reduces them to punk. 

Expansion of Blocks. Wood blocks expand on exposure to 
moisture; and, when they are laid end to end across the street, the 
curbstones are liable to be displaced, or the courses of the blocks will 
be bent into reserve curves. To avoid this, the joints of the courses 
near the curb may be left open until expansion has ceased, the space 
being temporarily filled with sand. The rate of expansion is about 
1 inch in 8 feet, but Caries for different woods. The time required for 
the wood to become fully expanded varies from 12 to 18 months. By 
employing blocks impregnated with the oil of creosote, this trouble will 
be avoided. 'Blocks so treated do not contract or expand to any appre- 
ciable extent. 

The comparative expansion of creosoted and plain wood blocks 
after immersion in water for forty-eight hours, in percentage on orig- 
inal dimensions, was: 

Expansion of Wood Paving Blocks 

On length of block.. 
Oil width " " . 
On depth " " . 

Manner of Laying. The blocks are set with the fiber vertical, 
and the long dimension crosswise of the street, the longitudinal joints 
being broken by a lap of at least one-third the length of the block; the 

blocks should be laid so as to have the least possible width of joint. 
Wide joints hasten the destruction of the wood by permitting- the fibers 
to wear under traffic, which also causes the surface of the pavement 



to wear in small ridges. The most recent practice for laying blocks 
on 3 per cent grades, has been to remove from the top of one side of 
each block a strip J inch thick and H inches deep, extending the length 
of the block. When the blocks are laid and driven closely together, 
there is a quarter-inch opening or joint extending clear across the street 
in each course. These joints are filled with Portland cement grout. 
Fig. 63 shows a section of pavement having this form of joint. 

Filling for Joints. The best materials for filling the joints are 
bitumen for the lower two or three inches, and hydraulic cement grout 
for the remainder of the depth. The cement grout protects the pitch 
from the action of the sun, and does not wear down very much below 
the surface of the wood. 


Asphaltic Paving Materials. All asphaltic or bituminous pave- 
ments are composed of two essential parts namely, the cementing 
material (matrix) and the resisting material (aggregate). Each has a 
distinct function to perform; the first furnishes and preserves the co- 
herency of the mass; the second resists the wear of traffic. 

Two classes of asphaltic paving compounds are in use, namely, 
natural and artificial. The "natural" variety is composed of either 
limestone or sandstone naturally cemented with bitumen. To this 
class belong the bituminous limestones of Europe, Texas, Utah, etc., 
and the bituminous sandstones of California, Kentucky, Texas, Indian 
Territory, etc. The "artificial" consists of mixtures of asphaltic 
cement with sand and stone dust. To this class belong the pavements 
made from Trinidad, Bermudez, Cuban, and similar asphaltums. 
For the artificial variety, most hard bitumens are, when properly 
prepared, equally suitable. For the aggregate, the most suitable mate- 
rials are stone-dust from the harder rocks, such as granite, trap, etc., 
and sharp angular sand. These materials should be entirely free from 
loam and vegetable impurities. The strength and enduring qualities 
of the mixture will depend upon the quality, strength, and proportion 
of each ingredient, as well as upon the cohesion of the matrix and its 
adhesion to the aggregate. 

Bituminous limestone consists of carbonate of lime naturally 
cemented with bitumen in proportions varying from 80 to 93 per cent 
of carbonate of lime and from 7 to 20 per cent of bitumen. Its color, 
when freshly broken, is a dark (almost black) chocolate brown, the 



darker color being due to a large percentage of bitumen. At a tem- 
perature of from 55 to 70 F., the material is hard and sonorous, and 
breaks easily with an irregular fracture; at temperatures between 70 
and 140 F. it softens, passing with the rise in temperature through 
various degrees of plasticity, until, at between 140 and 160 F., it 
begins to crumble; at 212 it commences to melt; and at 280 F. it is 
completely disintegrated. Its specific gravity is about 2.235. 

Bituminous limestone is the material employed for paving pur- 
poses throughout Europe. It is obtained principally from deposits 
at Val-de-Travers, canton of Neufchatel, Switzerland ; at Seysell, in 
the Department of Ain, France; at Ragusa, Sicily; at Limmer, near 
Hanover; and at Vorwohle, Germany. 

Bituminous limestone is found in several parts of the United 
States. Two of these deposits are at present being worked one in 
Texas, the material from which is called "lithocarbon"; and one on the 
Wasatch Indian Reservation. These deposits contain from 10 to 30 
per cent of bitumen. 

The bituminous limestones which contain about 10 per cent of 
bitumen are used for paving in their natural condition, being simply 
reduced to powder, heated until thoroughly softened, then spread while 
hot upon the foundation, and tamped and rammed until compacted. 

Bituminous sandstones are composed of sandstone rock impreg- 
nated with bitumen in amounts varying from a trace to 70 per cent. 
They are found in both Europe and America. In Europe, they are 
chiefly used for the production of pure bitumen, which is extracted by 
boiling or macerating them with water. In the United States, exten- 
sive deposits are found in the Western States; and since 1880 they have 
been gradually coming into use as a paving material, so that now up- 
wards of 150 miles of streets in Western cities are paved with them. 
They are prepared for use as paving material by crushing to powder, 
which is heated to about 250 F. or until it becomes plastic, then spread 
upon the street arid compressed by rolling; sometimes sand or gravel 
Js added, and it is stated that a mixture of about 80 per cent of gravel 
makes a durable pavement. 

Trinidad Asphaltum. The deposits of asphaltum in the island 
of Trinidad, W. I., have been the main source of supply for the asphal- 
tum used in street paving in the United States. Three kinds are found 
there, which have been named, according to the source, lake-pitch, 



land or overflow pitch, and iron pitch. The first and most valuable 
kind is obtained from the so-called Pitch Lake. 

The term land or overflow pitch is applied to the deposits of 
asphaltum found outside the lake. These deposits form extensive 
beds of variable thickness, and are covered with from a few to several 
feet of earth ; they are considered by some authorities to be formed from 
pitch which has overflowed from the lake; by others to be of entirely 
different origin. The name cheese pitch is given to such portions of the 
land pitch as more nearly resemble that obtained from the lake. 

The term iron pitch is used to designate large and isolated masses 
of extremely hard asphaltum found both within and without the bor- 
ders of the lake. It is supposed to have been formed by the action of 
heat caused by forest fires, which, sweeping over the softer pitch, re- 
moved its more volatile constituents. 

The name epuree is given to asphaltum refined on the island of 
Trinidad. The process is conducted in a very crude manner, in large, 
open, cast-iron sugar boilers. 

The characteristics of crude Trinidad asphaltum, both lake and 
land, are as follows: It is composed of bitumen mixed with fine sand, 
clay, arid vegetable matter. Its specific gravity varies according to the 
impurities present, but is usually about 1.28. Its color, when fleshly 
excavated, is a brown, which changes to black on exposure to the at- 
mosphere. When freshly broken, it emits the usual bituminous odor- 
It is porous, containing gas cavities, and in consistency resembles 
cheese. If left long enough in the sun, the surface will soften and melt, 
and will finally flow into a more or less compact mass. 

Refined Trinidad Asphaltum. The crude asphaltum is refined 
or purified by melting it in iron kettles or stills by the application of 
indirect heat. 

The operation of refining proceeds as follows : During the heat- 
ing, the water and lighter oils are evaporated; the asphaltum is lique- 
fied; the vegetable matter rises to the surface, and is skimmed off; the 
earthy and siliceous matters settle to the bottom; and the liquid asphal- 
tum is drawn off into old cement or flour barrels. 

When the asphaltum is refined without agitation, the residue 
remaining in the still forms a considerable percentage of the crude 
material, frequently amounting to 12 per cent; and it was at one time 
considered that the greater the amount of this residue the better the 




quality of the refined asphaltum. Since agitation has been adopted } 
however, the greater part of the earthy and siliceous matters is retained 
in suspension; and it has come to be considered just as desirable for 
a part of the surface mixture as the sand which is subsequently added. 
The refined asphaltum, if for local use, is generally converted into 
cement in the same still in which it was refined. 

The average composition of both the land and lake varieties is 
shown by the following analyses : 

Average Composition of Trinidad Asphaltum 







Per Cent 
26. H8 

Per Cent 

Per Cent 

Inorganic matter 
Organic non-bituminous matter 
Bitumen ... 

When the analyses are calculated to a basis of dry 
substances, the composition is : Inorganic matter 
Organic matter not bitumen 









The substances volatilized in 10 hours at 400 F . .. 
The substances soften at 
" " flow at 




190 F. 
200 F. 

170 F. 
185 F. 

0.86 to 1.37 
200 to 251)3 F. 
2 10 to 328 F. 

The characteristics of refined Trinidad asphaltum are as follows: 
The color is black, with a homogeneous appearance. At a tempera- 
ture of about 70 F., it is very brittle, and breaks with a conchoidal 
fracture. It burns with a yellowish-white flame, and in burning emits 
an empyreumatic odor, and possesses little cemsntitious quality. To 
give it the required plasticity and tenacity, it is mixed while liquid with 
from 16 to 21 pounds of residuum oil to 100 pounds of asphaltum. 

The product resulting from the combination is called asphalt 
paving-cement. Its consistency should be such that, at a temperature 
of from 70 to 80 F., it can be easily indented with the fingers, and on 
slight warming be drawn out in strings or threads. 

Artificial Asphalt Pavements. The pavements made from Trini- 
dad, Bermudez, California, and similar asphaltums, are composed of 
mechanical mixtures of asphaltic cement, sand, and stone-dust. 

The sand should be equal in quality to that used for hydraulic 
cement mortar; it must be entirely free from clay, loam, and vegetable 



impurities; its grains should be angular and range from coarse to fine. 

The stone-dust is used to aid in filling the voids in the sand and 
thus reduce the amount of cement. The amount used varies with the 
coarseness of the sand and the quality of the cement, and ranges from 
5 to 15 per cent. (The voids in sand vary from .3 to .5 per cent.) 

As to the quality of the stone-dust, that from any durable stone is 
equally suitable. Limestone-dust was originally used, and has never 
been entirely discarded. 

The paving composition is prepared by heating the mixed sand 
and stone-dust and the asphalt cement separately to a temperature of 
about 300 F. The heated ingredients are measured into a pu^-mill 
and thoroughly incorporated. When this is accomplished, the mix- 
ture is ready for use. It is hauled to the street and spread with iron 
rakes to such depth as will give the required thickness when compacted 
(the finished thickness varies between 1^ and 2j inches). The re- 
duction of thickness by compression is generally about 40 per cent. 

The mixture is sometimes laid in two layers. The first is called 
the binder or cushion=coat ; it contains from 2 to 5 per cent more cement 
than the surface-coat; its thickness is usually -i inch. The object of the 
binder course is to unite the surface mixture with the foundation, which 
it does through the larger percentage of cement that it contains, which, 
if put in the surface mixture, would render it too soft. 

The paving composition is compressed by means of rollers and 
tamping irons, the latter being heated in a fire contained in an iron 
basket mounted on wheels. These irons are used for tamping such 
portions as are inaccessible to the roller namely, gutters, around man- 
hole heads, etc. 

. Two rollers are sometimes employed; one, weighing 5 to 6 tons 
and of narrow tread, is used to give the first compression; and the 
other, weighing about 10 tons and of broad tread, is used for finishing. 
The amount of rolling varies; the average is about 1 hour per 1,000 
square yards of surface. After the primary compression, natural 
hydraulic or any impalpable mineral matter is sprinkled over the sur- 
face, to prevent the adhesion of the material to the roller and to give 
the surface a more pleasing appearance. When the asphalt is laid 
up to the curb, the surface of the portion forming the gutter is painted 
with a coat of hot cement. 

Although asphaltum is a bad conductor of heat, and the cement 





}M Qg 






retains its plasticity for several hours, occasions may and do arise 
through which the composition before it is spread has cooled; its con- 
dition when this happens is analogous to hydraulic cement which has 
taken a "set," and the same rules which apply to hydraulic cement in 
this condition should be respected in regard to asphaltic cement. 

The proportions of the ingredients in the paving mixture are not 
constant, but vary with the climate of the place where the pavement 
is to l>e used, the character of the sand, and the amount and character 
of the traffic that will use the pavement. The range in the proportion 
is as follows : 

Formula for Asphaltic Paving Mixture 

Asphalt cement 12 to 1.5 per cent. 

Sand 70 to 83 " " 

Stone-dust 5 to 15 " " 

A cubic yard of the prepared material weighs about 4,500 pounds, and 
will lay the following amount of wearing-surface: 

2i inches thick 12 square yards. 

2 " " 18 " 

H " " 27 " 

One ton of refined asphaltum makes about 2,300 pounds of asphalt 
cement, equal to about 3.4 cubic yards of surface material. 

Foundation. A solid, unyielding foundation is indispensable 
with all asphaltic pavements, because asphalt of itself has no power of 
offering resistance to the action of traffic, consequently it is nearh 
always placed upon a bed of hydraulic cement concrete. The concrete 
must be thoroughly set and its surface dry before the asphalt is laid 
upon it; if not, the water will be sucked up and converted into steam, 
with the result that coherence of the asphaltic mixture is prevented, 
and, although its surface may be smooth, the mass is really honey- 
combed, so that as soon as the pavement is subjected to the action of 
traffic, the voids or fissures formed by the steam appear on the surface, 
and the whole pavement is quickly broken up. 

Advantages of Asphalt Pavement. These may be summed up 
as follows: 

(1) Ease of traction. 

(2) It is comparatively noiseless under traffic. 

(3) It is impervious. 

(4) It is easily cleansed. 

(5) It produces neither mud nor dust. 

(6) It is pleasing to the eye. 



(7) It suits all classes of traffic. 

(8) There is neither vibration nor concussion in traveling over it. 

(9) It is expeditiously laid, thereby causing little inconvenience 
to traffic. 

(10) Openings to gain access to underground pipes are easily 

(11) It is durable. 

(12) It is easily repaired. 

Defects of Asphalt Pavement. These are as follows: 

(1) It is slippery under certain conditions of the atmosphere. 
The American asphalts are much less so than the European, on account 
of their granular texture derived from the sand. The difference is 
very noticeable ; the European are as smooth as glass, while the Ameri- 
can resemble fine sandpaper. 

(2) It will not stand constant moisture, and will disintegrate if 
excessively sprinkled. 

(3) Under extreme heat it is liable- to become so soft that it will 
roll of creep under traffic and present a wavy surface; and under ex- 
treme cold there is danger that the surface will crack and become 

(4) It is not adapted to grades steeper than 2-V per cent, although 
it is in use on grades up to 7.30 per cent. 

(5) Repairs must be quickly made, for the material has little 
coherence, and if, from irregular settlement of foundation or local vio- 
lence, a break occurs, the passing wheels rapidly shear off the sides of 
the hole, and it soon assumes formidable dimensions. 

The strewing of sand upon asphalt renders it less slippery; but in 
addition to the interference of the traffic while this is being done, there 
are further objections namely, the possible injury by the sand cutting 
into the asphalt, the expense of labor and materials, and the mud 
formed, which has afterwards to be removed. 

Although pure asphaltum is absolutely impervious and insoluble 
in either fresh or salt wafer, yet asphalt pavements in the continued 
presence of water are quickly disintegrated. Ordinary rain or daily 
sprinkling does not injure them when they are allowed to become per- 
fectly dry again. The damage is most apparent in gutters and adja- 
cent to overflowing drinking fountains. This defect has long been 
recognized; and various measures have been taken to overcome it, or 



at least to reduce it to a minimum. In some cities, ordinances have 
been passed, seeking to regulate the sprinkling of the streets; and in 
many places the gutters are laid with stone or vitrified brick (see Figs. 

Fig. 64. 

64 and 65), while in others the asphalt is laid to the curb, a space of 
12 to 15 inches along the curb being covered with a thin coating of 
asphalt cement. 

Asphalt laid adjoining center-bearing street-car rails is quickly 
broken down and destroyed. This defect is not peculiar to asphalt. 
All other materials when placed in similar positions are quickly worn. 
Granite blocks laid along such tracks have been cut into at a rate of 
more than half an inch a year. The frequent entering and turning off 
of vehicles from car tracks is one of the severest tests that can be 

j Asphalt 




Brick \ Gutter 


Fig. 65. 

applied to any paving material; moreover, the gauge of trucks and 
vehicles is frequently greater than that of the rails, so one wheel runs 
on the rail and the other outside. The number of wheels thus travel- 
ing in one line must quickly wear a rut in any material adjoining the 
center-bearing rail. 

To obviate the destruction of asphalt in such situations, it is usual 
to lay a strip of granite block or brick paving along the rail. This 
pavement should be of sufficient width to support the wheels of the 
widest gauge using the street. 



The burning of leaves or making of fires on asphalt pavements 
should not be permitted, as it injures the asphalt, and the paving com- 
panies cannot be compelled to repair the damaged places without 

Asphalt Blocks. Asphalt paving blocks are formed from a mix- 
ture of asphaltic cement and crushed stone in the proportion of 8 to 12 
per cent of cement to 88 and 92 per cent of stone. The materials are 
heated to a temperature of about 300 F., and mixed while hot in a 
suitable vessel. When the mixing is complete, the material is placed 
in moulds and subjected to heavy pressure, after which the blocks are 
cooled suddenly by plunging into cold water. 

The usual dimensions of the blocks are 4 inches wide, 3 inches 
deep, and 12 inches long. 

Foundation. The blocks are usually laid upon a concrete founda- 
tion with a cushion-coat of sand about \ inch thick. They are laid 
with their length at right angles to the axis of the street, and the longitu- 
dinal joints should be broken by a lap of at least 4 inches. The blocks 
are then either 'rammed with hand rammers 01 rolled with a light steam 
roller, the surface being covered with clean, fine sand ; no joint filling is 
used, as, under the action of the sun and traffic, the blocks soon become 

The advantages claimed for a pavement of asphalt blocks over a 
continuous sheet of asphalt are: (1) That they can be made at a 
factory located near the materials, whence they can be transported to 
the place where they are to be used and can be laid by ordinary paviors, 
whereas sheet pavements require special machinery and skilled labor; 
^2) that they are less slippery, owing to the joints and the rougher 
surface due to the use of crushed stone. 

Asphalt Macadam Bituminous Macadam. Recently it has been 
proposed to use asphalt as a binding material for broken stone. 
There are two patented processes the Whinery and the Warren 
which differ slightly in details. 

The advantages claimed for these methods are: (1) The first 
coat will be materially less; (2) it will offer a better foothold for horses; 
(3) it will be at least as durable as the ordinary sheet asphalt ; (4) it will 
not shift under traffic and roll into waves; (5) it will not crack; (6) it 

g 5 




can be repaired more cheaply and with less skilled labor than can the 
ordinary sheet asphalt. 

Tools Employed in Construction of Asphalt Pavements. The 

*:*-~s -..- 


Fig. 66. Steam Holler. 

FiK- 67. Asphalt Tools 

tools used in laying sheet asphalt pavements comprise iron rakes; 
hand rammers; smoothing irons (Fig. 67); pouring pots (Fig. 69): 




hand rollers, either with or without a fire-pot (Fig. 68); and steam 
rollers, with or without provision for heating the front roll (Fig. 66). 
These rollers are different in construction, appearance, and weight 

Fig. 68. Hand Rollers. 

Fig. 69. Pouring Pots. 

from those employed for compacting broken stone. The difference 
is due to the different character of the work recjuired. 

The principal dimensions of a five-ton roller are as follows : 

Front roll or steering-wheel 30 to 32 inches diameter. 

Rear roll or driving-wheel 48 

Width of front roll 40 

" rear " 40 " 

Extreme length 14 feet. 

height 7 to 8 feet. 

Water capacity 80 to 100 gallons. 

Coal " 200 pounds. 



A footpath or walk is simply a road under another name a road 
for pedestrians instead of one for horses and vehicles. The only 
difference that exists is in the degree of service required; but the con- 
ditions of consruction that render a road well adapted to its object are 
very much the same as those required for a walk. 

The effects of heavy loads such as use carriageways are not felt 
upon footpaths; but the destructive action of water and frost is the 
same in either case, and the treatment to counteract or resist these 
elements as far as practicable, and to produce permanency, must be 
the controlling idea in each case, and should be carried out upon a 
common principle. It is not less essential that a walk should be well 
adapted to its object than that a road should be; and it is annoying to 
find it impassable or insecure and in want of repair when it is needed 
for convenience or pleasure. In point of economy, there is the same 
advantage in constructing a footway skilfully and durably as there is 
in the case of a road. 

Width. The width of footwalks (exclusive of the space occupied 
by projections and shade trees) should be ample to accomodate com- 
fortably the number of people using them. In streets devoted entirely 
to commercial purposes, the clear width should be at least one-third 
the width of the carriageway; in residential and suburban streets, a 
very pleasing result can be obtained by making the walk one-half the 
width of the roadway, and devoting the greater part to grass and shade 

Cross Slope. The surface of footpaths must be sloped so that 
the surface water will readily flow to the gutters. This slope need not 
be very great; inch per foot will be sufficient. A greater slope with a 
thin coating of ice upon it, becomes dangerous to pedestrians. 

Foundation. As in the case of roadways, so with footpaths, the 
foundation is of primary importance. Whatever material may be used 
for the surface, if the foundation is weak and yielding, the surface will 
settle irregularly and become extremely objectionable, if not danger- 
ous, to pedestrians. 

Surface. The requirements of a good covering for sidewalks are : 

(1) It must be smooth but not slippery. 

(2) It must absorb the minimum amount of water, so that it 
may dry rapidly after rain. 



(3) It must not be easily abraded. 

(4) It must be of uniform quality throughout, so that it may 
wear evenly. 

(5) It must neither scale nor flake. 

(6) Its texture must be such that dust will not adhere to it. 

(7) It must be durable. 

Materials. The materials used for footpaths are as follows: 
Stone, natural and artificial; wood; asphalt; brick; tar concrete; and 

Of the natural stones, sandstone (Milestone) and granite are ex- 
tensively employed. 

The bluestone, when well laid, forms an excellent paving material. 
It is of compact texture, absorbs water to a very limited extent, and 
hence soon dries after rain ; it has sufficient hardness to resist abrasion, 
and wears well without becoming excessively slippery. 

Granite, although exceedingly durable, w r ears very slippery, and 
its surface has to be frequently roughened. 

Slabs, of whatever stone, must be of equal thickness throughout 

their entire area; the 
edges must be dressed 
true to the square for the 
whole thickness (edges 
must not be left feather- 
ed as shown in Fig. 70) ; and the slabs must be solidly bedded on the 
foundation and the joints filled with cement-mortar. 

Badly set or faultily dressed flagstones are very unpleasant to 
walk over, especially in rainy weather; the unevenness causes pedes- 
trians to stumble, and rocking stones squirt dirty w r ater over their 

Wood has been largely used in the form of planks ; it is cheap in 
first cost, but proves very expensive from the fact that it lasts but a 
comparatively short time and requires constant repair to keep it 
from becoming dangerous. 

Asphalt forms an excellent footway pavement; it is durable and 
does not wear slippery. 

Brick. Brick of suitable quality, well and carefully laid on a 
concrete foundation, makes an excellent footway pavement for resi- 


dential and suburban streets of large cities, and also for the main 
streets of smaller towns. The bricks should be a good quality of 
paving brick (ordinary building brick are unsuitable, as they soon 
wear out and are easily broken) . The bricks should be laid in parallel 
rows on their edges, with their length at right angles to the axis of the 

Curbstones. Curbstones are employed for the outer side of foot- 
ways, to sustain the coverings and form the gutter. Their upper edges 
are set flush with the foot walk pavement, so that the water' can flov, 
over them into the gutters. 

The disturbing forces which the curb has to resist, are : (1) The 
pressure of the earth behind it, which is frequently augmented by 
piles of merchandise, building materials, etc. This pressure tends to 
overturn it, break it transversely, or move it bodily on its base. (2) 
The pressure due to the expansion of freezing earth behind and be- 
neath it. This force is most frequent where the sidewalk is partly 
sodded and the ground is accordingly moist. Successive freezing and 
thawing of the earth behind the curb will occasion a succession of 
thrusts forward, which, if the curb be of faulty design, will cause it. to 
incline several degrees from the vertical. (3) The concussions and 
abrasions caused by traffic To withstand the destructive effect of 
wheels, curbs are faced with iron; and a concrete curb with a rounded 
edge of steel has been patented and used to some extent. Fires built 
in the gutters deface and seriously injure the curb. Posts and trees 
set too near the curb, tend to break, displace, and destroy it. 

The use of drain tiles under the curb is a subject of much differ- 
ence of opinion among engineers. Where the subsoil contains water 
naturally, or is likely to receive it from outside the curb-lines, the use 
of drains is of decided benefit; but great care must be exercised in 
jointing the drain-tiles, lest the soil shall be loosened and removed, 
causing the curb to drop out of alignment. 

The materials employed for curbing are the natural stones, as 
granite, sandstone (bluestone), etc., artificial stone, fire-clay, and cast 

The dimensions of curbstones vary considerably in different 
localities and according to the width of the footpaths; the wider the 
path, the wider should be the curb. It should, however, never be less 
than 8 inches deep, nor narrower than 4 inches. Depth is necessary 




to prevent the curb turning over toward the gutter. It should never 
be in smaller lengths than 3 feet. The top surface should be beveled 
off to conform to the slope of the footpath. The front face should be 
hammer-dressed for a depth of about 6 inches, in order that there may 
be a smooth surface visible against the gutter. The back for 3 inches 

Fig. 71. 

from the top should also be dressed, so that the flagging or other paving 
may butt fair against it. The end joints should be cut truly square, 
the full thickness of the stone at the top, and so much below the top as 
will be exposed , the remaining portion of the depth and bottom should 
be roughly squared, and the bottom should be fairly parallel to the 
top. (See Figs. 71 and 72). 

Artificial Stone. Artificial stone is being extensively used as a 

footway paving material. Its manufacture is the subject of several 
patents, and numerous kinds are to be had in the market. When 
manufactured of first-class materials and laid in a substantial manner, 
with proper provision against the action of frost, artificial stone forms 
a durable, agreeable, and inexpensive pavement. 



Fig. 73. Tamper. 

The varieties most extensively used in the United States* are 
known by the names of granolithic, monolithic, ferrolithic, kosmocrete, 
metalithic, etc. 

The process of manufacture is practically the same for all kinds, 
the difference being in the materials em- 
ployed. The usual ingredients are Port- 
land cement, sand, gravel, and crushed 

Artificial stone for footway pave- 
ments is formed in two ways namely, 
in blocks manufactured at a factory, 
brought on the ground, and laid in the 
same manner as natural stone; or the raw 
materials are brought upon the work, pre- 
pared, and laid in place, blocks being formed by the use of board 

The manner of laying is practically the same for all kinds. The 
area to be paved is excavated to a mini- 
mum depth of 8 inches, and to such great- 
er depths as the nature of the ground may 
require to secure a solid foundation. The 
surface of the ground so exposed is well 
compacted by ramming; and a layer of 
gravel, ashes, clinker, or other suitable material is spread and consoli- 
dated; on this is placed the concrete wearing surface, usually 4 inches 
thick. As a protection against the lifting 
effects of frost, the concrete is laid in 
squares, rectangles, or other forms hav- 
ing areas ranging from 6 to 30 square 
feet, strips of wood being employed to 
form moulds in which the concrete is 

placed. After the concrete is set, these strips are removed, leaving 
joints about half an inch wide between the blocks. Under some 
patents these joints are filled with cement; under others, with tarred 
paper; and in some cases they are left open. 

Tools Employed in Construction of Artificial Stone Pavements. 
Tampers (Fig. 73). Cast iron, with hickory handle; range from 6 
by 8 inches to 8 by 10 inches. 

Fig. 74. Quarter-Round. 




Quarter-Round, (Fig. 74). 
for forming corners and edges. 

Fig. 7ti. Cutter. 

Made of any desired radius. Used 

Jointer (Fig. 75). Used for 
trimming and finishing the joints. 

Cutter (Fig. 76). Used for cut- 
ting the concrete into blocks. 

Gutter Tool (Fig. 77). Used 
for forming and finishing gutters. 

Imprint Rollers (Figs. 78 and 
79). Here are shown two designs 
of rollers for imprinting the surface 
of artificial stone pavements with 
grooves, etc. 


The problem of selecting the best pavement for any particular 
case is a local one, not only for each city, but also for each of the various 
parts into which the city is imperceptibly divided; and it involves so 
many elements that the nicest balancing of the relative values for each 
kind of pavement is required, to arrive at a correct conclusion. 

In some localities, the proximity of one or more paving materials 
determines the character of the pavement ; 
while in other cases a careful investigation 
may be required in order to select the 
most suitable material. Local conditions 
should always be considered; hence it is 
not possible to lay down any fixed rule as to what material makes the 
best pavement. 

The qualities essential to a good pavement may be stated as 
follows : 

(1) It should be impervious. 

(2) It should afford good foothold for horses. 

(3) It should be hard and durable, so as to resist wear and dis- 

(4) It should be adapted to every grade. 

(5) It should suit every class of traffic. 

(6) It should offer the minimum resistance to traction. 

(7) It should be noiseless. 

(8) It should yield neither dust nor mud. 





(9) It should be easily cleaned. 

(10) It should be cheap. 

Interests Affected in Selection. Of the above requirements, 
numbers 2, 4, 5, and affect the traffic and determine the cost of haul- 
age by the limitations of loads, speed, 
and wear and tear of horses and ve- 
hicles. If the surface is rough or the 
foothold bad, the weight of the load 
a horse can draw is decreased, thus ne- 
cessitating the making of more trips or 
the employment of more horses and 
vehicles to move a given weight. A 
defective surface necessitates a reduc- 
tion in the speed of movement and 
consequent loss of time; it increases the 
w r ear of horses, thus decreasing their 
Fig. ra. imprint Roller. jjf e serv i ce and lessening the value of 

their current services ; it also increases 
the cost of maintaining vehicles and 

Numbers 7, 8, and 9 affect the 
occupiers of adjacent premises, who 
suffer from the effect of dust and 
noise; they also affect the owners of 
said premises, whose income from 
rents is diminished where these disad- 
vantages exist. Numbers 3 and 10 af- 
fect the taxpayers alone first, as to 
the length of time during which the 
covering remains serviceable; and sec- 
ond, as to the amount of the annual 

repairs. Number 1 affects the adjacent occupiers principally on 
hygienic grounds. Numbers 7 and 8 affect both traffic and occupiers. 
Problem Involved in Selection. The problem involved in the 
selection of the most suitable pavement consists of the following 
factors: (1) adaptability; (2) desirability; (3) serviceability; (4) dura- 
bility; (5) cost. 

Adaptability. The best pavement for any given roadway will 



depend altogether on local circumstances. Pavements must be adapt- 
ed to the class of traffic that will use them. The pavement suitable 
for a road through an agricultural district will not be suitable for the 
streets of a manufacturing center; nor will the covering suitable for 
heavy traffic be suitable for a pleasure drive or for a residential district. 

General experience indicates the relative fitness of the several 
materials as follows: 

For country roads, suburban streets, and pleasure drives broken 
stone. For streets having heavy and constant traffic rectangular 
blocks of stone, laid on a concrete foundation, with the joints filled 
with bituminous or Portland cement grout. For streets devoted to 
retail trade, and where comparative noiselessness is essential asphalt, 
wood, or brick. 

Desirability, The desirability of a pavement is its possession of 
qualities which make it satisfactory to the people using and seeing it. 
Between two pavements alike in cost and durability, people will have 
preferences arising from the condition of their health, presonal pre- 
judices, and various other intangible influences, causing them to select 
one rather than the other in their respective streets. Such selections 
are often made against the demonstrated economies of the case, and 
usually in ignorance of them. Whenever one kind of pavement is 
more economical and satisfactory to use than is any other, there should 
not be any difference of opinion about securing it, either as a new 
pavement or in the replacement of an old one. 

The economic desirability of pavements is governed by the ease 
of movement over them, and is measured by the number of horses or 
pounds of tractive force required to move a given weight usually one 
ton over them. The resistance offered to traction by different pave- 
ments is shown in the following table: 

Resistance to Traction on Different Pavements 



Pounds per ton 

In terms oftheUxid 

Asphalt (sheet) 


to 70 

cV to 




" 40 

Tsi " 




" 100 

Stone-block . " . 


" 80 


Wood-block rectangular 


" 50 

eV " 


Wood-block round 


" 80 





Serviceability. The serviceability of a pavement is its quality of 
fitness for use. This quality is measured by the expense caused to the 
traffic using it namely, the wear and tear of horses and vehicles, loss 
of time, etc. No statistics are available from which to deduce the 
actual cost of wear and tear. 

The serviceability of any pavement depends in great measure 
upon the amount of foothold afforded by it to the horses provided, 
however, that its surface be not so rough as to absorb too large a per- 
centage of the tractive energy required to move a given load over it. 
Cobblestones afford excellent foothold, and for this reason are largely 
employed by horse-car companies for paving between the rails; but 
the resistance of their surface to motion requires the expenditure ot 
about 40 pounds of tractive energy to move a load of 1 ton. Asphalt 
affords the least foothold; but the tractive force required to overcome 
the resistance it offers to motion is only about 30 pounds per ton. 

Comparative Safety. The comparison of pavements in respect of 
safety, is the average distance traveled before a horse falls. The 
materials affording the best foothold for horses are as follows, stated 
in the order of their merit : 

(1) Earth, dry and compact. 

(2) Gravel. 

(3) Broken stone (macadam). 

(4) Wood. 

(5) Sandstone and brick. 
(G) Asphalt. 

(7) Granite blocks. 

Durability. The durability of pavement is that quality which 
determines the length of time during which it is serviceable, and does 
not relate to the length of time it has been down. The only measure 
of durability of a pavement if the amount of traffic tonnage it will bear 
before it becomes so worn that the cost of replacing it is less than the 
expense incurred by its use. 

As a pavement is a construction, it necessarily follows that there 
is a vast difference between the durability of the pavement and the 
durability of the materials of which it is made. Iron is eminently 
durable; but, as a paving material, it is a failure. 

Durability and Dirt. The durability of a paving material will 
vary considerably with the condition of cleanliness observed. One 



inch of overlying dirt will most effectually protect the pavement from 
abrasion, and indefinitely prolong its life. But the dirt is expensive, 
it injures apparel and merchandise, and is the cause of sickness and 
discomfort. In the comparison of different pavements, no traffic 
should he credited to the dirty one. 

Life of Pavements. The life or durability of different pavements 
under like conditions of traffic and maintenance, may be taken as 
follows : 

Life Terms of Various Pavements 




Granite block 



to no years 



' 14 



3 ' 





. ? , 

Cost. The question of cost is the one which usually interests 
taxpayers, and is probably the greatest stumbling-block in the attain- 
ment of good roadways. The first cost is usually charged against the 
property abutting on the highway to be improved. The result is that 
the average property owner is always anxious for a pavement that costs 
little, because he must pay for it, not caring for the fact that cheap 
pavements soon wear out and become a source of endless annoyance 
and expense. Thus false ideas of economy always have stood, and 
undoubtedly to some extent always will stand, in the way of realizing 
that the best is the cheapest. 

The pavement which has cost the most is not always the best; nor 
is that which cost the least the cheapest ; the one which is truly the cheap- 
est is the one which makes the most profitable returns in proportion to the 
amount expended upon it. No doubt there is a limit of cost to go be- 
yond which would produce no practical benefit ; but it will always be 
found more economical to spend enough to secure the best results, and 
this will always cost less in the long run. One dollar well spent is 
many times more effective than one-half the amount injudiciously 
expended in the hopeless effort to reach sufficiently good results. The 
cheaper work may look as well as the more expensive for the 
time, but may very soon have to be done over again. 

Economical Benefit. The economic benefit of a good roadway is 


comprised in its cheaper maintenance; the greater facility it offers for 
traveling, thus reducing the cost of transportation; the lower cost of 
repairs to vehicles, and less wear of horses, thus increasing their term 
of serviceability and enhancing the value of their present service; the 
saving of time; and the ease and comfort afforded to those using the 

First Cost. The cost of construction is largely controlled by the 
locality of the place, its proximity to the particular material used, and 
the character of the foundation. 

The Relative Economies of Pavements whether of the same 
kind in different condition, or of different kinds in like good condition 
are sufficiently determined by summing their cost under the following 
headings of account: 

(1) Annual interest upon first cost. 

(2) Annual expense for maintenance. 

(3) Annual cost for cleaning and sprinkling. 

(4) Annual cost for service and use. 

(5) Annual cost for consequential damages. 

Interest on First Cost. The first cost of a pavement, like any 
other permanent investment, is measurable for purposes of comparison 
by the amount of annual interest on the sum expended. Thus, assum- 
ing the worth of money to be 4%, a pavement costing $4 per square 
yard entails an annual interest loss or tax of $0.16 per square yard. 

Cost of Maintenance. Under this head must be included all out- 
lays for repairs and renewals which are made from the time when the 
pavement is new and at its best to a time subsequent, when, by any 
treatment, it is again put in equally good condition. The gross sum 
so derived, divided by the number of years which elapse between the 
two dates, gives an average annual cost for maintenance. 

Maintenance means the keeping of the pavement in a condition 
practically as good as when first laid. The cost will vary considerably 
depending not only upon the material and the manner in which it is 
constructed, but upon the condition of cleanliness observed, and the ' 
quantity and quality of the traffic using the pavement. 

The prevailing opinion that no pavement is a good one unless, 
when once laid, it will take care of itself, is erroneous; there is no such 
pavement. All pavements are being constantly worn by traffic and 
by the action of the atmosphere; and if any defects which appear are 



not quickly repaired, the pavements soon become unsatisfactory and 
are destroyed. To keep them in good repair, incessant attention is 
necessary, and is consistent with economy. Yet claims are made that 
particular pavements cost little or nothing for repairs, simply because 
repairs in these cases are not made, while any one can see the need of 

Cost of Cleaning and Sprinkling. Any pavement, to be con- 
sidered as properly cared for, must be kept dustless and clean. While 
circumstances legitimately determine in many cases that streets must 
be cleaned at daily, weekly, or semi-weekly intervals, the only admis- 
sible condition for the purpose of analysis of street expenses must be 
that of like requirements in both or all cases subjected to comparison. 

The cleaning of pavements, as regards both efficiency and cost, 
depends (1) upon the character of the surface; (2) upon the nature of 
the materials of which the pavements are composed. Block pave- 
ments present the greatest difficulty; the joints can never be perfectly 
cleaned. The order of merit as regards facility of cleansing, is: (1) 
asphalt, (2) brick, (3) stone, (4) wood, (5) macadam. 

Cost of Service and Use. The annual cost for service is made up 
by combining several items of cost incidental to the use of the pave- 
ment for traffic for instance, the limitation of the speed of movement, 
as in cases where a bad pavement causes slow driving and consequent 
loss of time; or cases where the condition of a pavement limits the 
weight of the load which a horse can haul, and so compels the making 
of more trips or the employment of more horses and vehicles ; or cases 
where conditions are such as to cause greater wear and tear of vehicles, 
of equipage, and of horses. If a vehicle is run 1,500 miles in a year, 
and its maintenance costs $30 a year, then the cost of its maintenance 
per mile traveled is two cents. If the value of a team's time is, say, 
$1 for the legitimate time taken in going one mile with a load, and in 
consequence of bad roads it takes double that time, then the cost to 
traffic from having to use that one mile of bad roadway is $1 for each 
load. The same reasoning applies to circumstances where the weight 
of the load has to be reduced so as to necessitate the making of more 
than one trip. Again, bad pavements lessen not only the life-service 
of horses, but also the value of their current service. 

Cost for Consequential Damages. The determination of conse- 
quential damages arising from the use of defective or unsuitable pave- 


meats, involves the consideration of a wide array of diverse circum- 
stances. Hough-surfaced pavements, when in their best condition, 
afford a lodgment for organic matter composed largely of the urine 
and excrement of the animals employed upon the roadway. In 
warm and damp weather, these matters undergo putrefactive fer- 
mentation, and become the most efficient agency for generating and 
disseminating noxious vapors and disease germs, now recognized as 
the cause of a large part of the ills afflicting mankind. Pavements 
formed of porous materials are objectionable on the same, if not even 
stronger, grounds. 

Pavements productive of dust and mud are objectionable, and 
especially so on streets devoted to retail trade. If this particular 
disadvantage be appraised at so small a sum per lineal foot of frontage 
as SI. 50 per month, or six cents per day, it exceeds the cost of the best 
quality of pavement free from these disadvantages. 

Rough -surfaced pavements are noisy under traffic and insufferable 
to nervous invalids, and much nervous sickness is attributable to them. 
To all persons interested in nervous invalids, this damage from noisy 
pavements is rated as being far greater than would be the cost of sub- 
stituting the best quality of noiseless pavement; but there are, under 
many circumstances, specific financial losses, measurable in dollars 
and cents, dependent upon the use of rough, noisy pavements. They 
reduce the rental value of buildings and offices situated upon streets 
so paved offices devoted to pursuits wherein exhausting brain work 
is required. In such locations, quietness is almost indispensable, 
and no question about the cost of a noiseless pavement weighs against 
its possession. When an investigator has done the best he can to 
determine such a summary of costs of a pavement, he may divide the 
amount of annual tonnage of the street traffic by the amount of annual 
costs, and know what number of tons of traffic are borne for each cent 
of the average annual cost, which is the crucial test for any comparison, 
us follows: 

(1) Annual interest upon first cost $ 

(2) Average annual expense for maintenance and renewal . . . 

(3) Annual cost for custody (sprinkling and cleaning) 

(4) Annual cost for service and use 

(5) Annual cost for consequential damages 

Amount of average annual cost 

Annual tonnage of traffic 

Tons of traffic for each cent of cost 

Gross Cost of Pavements. Since the cost of a pavement depends 



upon the material of which it is formed, the width of the roadway, the 
extent and nature of the traffic, and the condition of repair and clean- 
liness in which it is maintained, it follows that in no two streets is the 
endurance or the cost the same, and the difference between the highest 
and lowest periods of endurance and- amount of cost is very con- 

The comparative cost of the various street pavements, including 
interest on first cost, sinking fund, maintenance, and cleaning, when 
reduced to a uniform standard traffic of 100,000 tons per annum for 
each yard in width of the carriageway, is about as follows: 
Comparative Cost of Various Pavements 



Granite blocks $0.25 


Asphalt street ... 0.40 


Wood ..., 



In the foregoing sections of this Cyclopedia nu- 
merous illustrative examples are worked out in 
detail in order to show the application of the 
various methods and principles. Accompanying 
these are examples for practice which will aid the 
reader in fixing the principles in mind. 

In the following pages are given a larg. num- 
ber of test questions and problems which afford a 
valuable means of testing the reader's knowledge 
of the subjects treated. They will be found excel- 
lent practice for those preparing for Civil Service 
Examinations. In some cases numerical answers 
are given as a further aid in this work. 






1. Write a short history of early bridges. 

2. Define: Truss, bridge truss, truss bridge, girders, and 
girder bridges. 

3. Draw an outline of a through bridge, and also an outline of 
a deck bridge. 

4. Make an outline diagram of a truss, and write the names 
of the various parts on the respective members. 

5. Make an outline diagram of a Warren, Howe, Pratt, 
bowstring, and Baltimore truss. 

6. Compute the weight of steel in a 130-foot highway bridge 
whose trusses are 16 feet center to center, given W = 34 + 22& + 
0.1GW + 0.7/. 

7. Compute the weight of steel in a deck plate-girder span of 
100 feet. Loading, E 50. Given W - 124.0 + 12. 01. 

8. What are equivalent uniform loads? 

9. What is Cooper's E 40 loading? 

10. Prove that the stress in a diagonal of a horizontal chord 
truss with a simple web system is V sec <. 

1 1 . Prove that the chord stress is M -=- h, where M is the moment 
at the point, and h is the height of the truss. 

12. Prove that the load must be on the segment of the span to 
the right of the section to produce the maximum positive shear. 

13. Compute the maximum positive and negative live-load 
shears in a 13-panel Howe truss, the live panel load being 40 000 



T H K H U B J K tJ T OK 



1 . Write an essay of 200 words on the economic considerations 
governing the decision to build and the decision as to what kind of 
bridge to employ. 

2. What determines the height and width of railroad truss 

3. Draw a clearance diagram for a bridge on a straight track, 
and state what allowance should be made if the bridge is on a curve. 

4. Describe a stress sheet, and tell what should be on it. 

5. Make a sketch of a cross-section of a deck plate-girder, 
showing the cross-ties, guard-rails, and rails in place. 

6. Make a sketch showing how tracks on curves are con- 

7. WTiat is the span under coping, the span center to center of 
bearings, and the span over all? 

8. Design a tie for Cooper's E 50 loading. 

9. If the end shear of a plate-girder is 394 500 pounds, design 
the web section, it being 108 inches deep. 

10- If the dead-load moment is 8 489 000 pound-inches and 
the live-load moment is 30 610 000 pound-inches, design the flange, 
if the distance back to back of flange angle, is 7 feet 6^ inches, it being 
assumed that the web does not take any bending moment. 

11. If, in the girder of Question 10, above, the web were 90 by 
-j^-inch, design the flange, considering ^ of the gross area of the web 
as effective flange area. 



O N T H K SU H,l K C T O F 


1 . Upon what does the ease with which a vehicle can he moved 
on a road depend? 

2. What kind of a road surface offers the greatest resistance to 

3. How may the power required to draw a vehicle over a pro- 
jecting stone be calculated? 

4. What effect has gravity on the load a horse can pull? 

5. Under what condition is the tractive power of a horse de- 

6. What are the best methods for improving sand roads? 

7 State briefly how earth is loosened and transported and the 
conditions under which each method is most advantageous? 

8. What are the essential requisites for securing a good gravel 

9. How should gravel roads be repaired? 

10. State the considerations that control the maximum grade. 

11. How should different grades be joined? 

12. What considerations control the width of a road? 

13. What is the essential quality of a stone used for road 

14. What should be the shape and size of broken stone? 

15. For a light traffic road what thickness should the layer of 
broken stone have? 

16. How should the foundation for the broken stone be pre- 





1. How should the natural soil be prepared to receive a 
pavement ? 

2. In ramming blocks in the pavement, what point requires 
to be watched ? 

3. How is a sand cushion prepared for use ? 

4. What influences the durability of a granite? 

5. How are rectangular stones laid on steep grades ? 

6. How is the surface and sub-surface drainage of streets 
provided for? 

7. W hat are the principal objections to wood pavements? 

8. What determines the best width for a street? 

9. In filling the joints with gravel and bituminous cement, 
what should be the condition of the material? 

10. What controls the maximum grade for a given street? 

11. What varieties of wood give the most satisfactory results? 

12. To what tests are stones intended for paving subjected? 

13. Do cobblestones form a satisfactory pavement? 

14. What properties should a stone possess to produce a sat- 
isfactory paving block? 

15. How are expansion joints formed in a pavement? 

16. What is the most suitable material for the foundation of 
a pavement? 

17. Under what class of traffic may wood be used? 

18. Upon what does the durability of a pavement depend? 

19. What materials are employed for filling the joints be- 
tween the paving blocks? 



The page numbers of this volume ivill be found at the bottom of the 
pages; the numbers at the top refer only to the section. 




Bridge Engineering 

Asphalt pavement 




advantages of 


bridge design 


defects of 


clearance diagram 




economic considerations 


materials for 


economic proportions 


tools used in construction of 


floor system 


Asphaltic paving mixture, formula for 


practical considerations 


Axle friction 





stress sheet 


weights and loadings 


Baltimore truss 




Belgian block pavement 
Bituminous limestone 


Bridge trusses, definition of 
Bridges, loads for 



Bowstring truss 


live loads 


Brick pavements 


wind loads 


absorption test 


Bridges, weights of 


advantages of 


formula; for 


cross-breaking test 


highway spans 


crushing test 


railroad spans 


defects of 


Broken-stone roads 


foundation for 


rattler test 



Bridge analysis 




early bridges 


City streets 


truss bridge development 


arrangement of 


Bridge engineering 11-264 

asphalt pavement 




Belgian block pavement 


bridge analysis 


brick pavements 








cobblestone pavement 


girder spans 




















trusses 16 

i, 53 

granite block pavement 


Note. For page numbers see foot of pages. 






City streets 




Deck bridges, definition of 


stone-block pavements 


Drainage of ro;ids 


1 ransverse contour 


Drains, fall of 


width of 


wood pavements 



Cobblestone pavements 






Earth roads 


Concrete-mixing machine 


Earthenware pipe culverts 


Country roads 






balancing cuts and fills 




classilicat ion of 

3 1 3 

fall of drains 


embankments on hillsides 


side ditches 


prosecution of 


of surface 


shrinkage of 

3 1 2 

water breaks 






Embankments on hillsides 


general considerations 

Engine loads in computing stresses 


axle friction. 


maximum moments, position of wheel 

efTect of springs on vehicles 


loads for 




maximum shear, position of wheel 

loss of tractive i>ower on inclines 274 

loads for 


object of roads 



resistance of air 


Final stresses 

1 1 

resistance to rolling 


Floor-l>eams, moments and shears in 


tractive i>ower and gradients 


Floor systems 


location of 




bridge sites 


Forces, resolution of 

cross levels 





final selection 


Girder bridges, definition of 


intermediate towns 


Girder spans 




moments in plate-girders 

1 19 



moments and shears in floor-l>eam 




shears in plate-girders 




stresses in plate-girders 




Girders, definition of 




( ; rades 

road coverings 




transverse contour 
















Grading tools 






materials for 






dump cars 


Note. For page numbers see foot of pages 






Grading tools 

Mountain roads 


dump wagons 




horse rollers 


construction profile 


mechanical graders 


establishing grade 




final location 






road machine 


halting places 




level stretches 




loss of height 


sprinkling carts 


maximum grade 

29 1 

surface graders 


minimum grade 




undulating grades 


Granite block pavement, 


water on 


Gravel heaters 




Gravel roads 





Parabolic truss 





Hard pan 




Highway bridges, live loads for 


Belgian block 


Highway construction . 267-398 



city streets 




country roads 


concrete foundation for 








granite block 


grading tools 




mountain roads 


stone block 






road coverings 


Plate-girder .railway-span design 

1 56 

Highway spans, actual weights of 




Howe truss 





determination of class 

1 56 

Inclines, loss of tractive power on 


determination of span 


Iron pipe culverts 




lateral systems 



masonry plan 


Knee-braces, definition of 



1 92 


stress sheet 


Lateral bracing, definition of 
Live load, position of, for maximum 


ties- and guard-rails 
web, economic depth of 
web splice 

1 59 

positive and negative shears 
Ix)os3 rock 


moments in 

1 19 


shears in 


Maximum and minimum stresses 


stresses in 


Mechanical graders 


Pony-truss bridge, definition of 


Melting furnaces 


Portals, definition of 


Moments in plate-girders 


Pratt truss 

48, 53, 93 

Note. For page numbers see foot of pages 








Railroad spans, actual weights of 


Railway bridges, live loads for 


Road coverings 


332 Tabl 



Road machines 





drainage of 








Roadways on rock slopes 



Sand roads 


Shears, maximum live-load 


Shears in plate-girders 


Shoes and roller nests 


Side slopes 




covering of 


form of 


inclination of 


Snow-load stresses 


Solid rock 




Stone block pavements 


Street grades 







Stresses in bridge trusses 


in chord members 




engine loads 


live-load moments 


live-load shears 


maximum and minimum 


moments method 


notation used 

35 1 

resolution of forces method 

27 1 

Warren truss under dead load 

35 1 

Warren truss under live load 


in web members 

31 i 

Stresses in plate-girders 


Note. For page numbers see foot of 


Surface graders 

Sway bracing, definition of 




bridges, formulae for weights of 19 

bridges, types of for various spans 142 
dead-load chord stresses 53, 61 

dead-load stresses in diagonals 54 

deck plate-girders, weights of 21 

floor-beam reactions 119 
force required to draw loaded vehicles 

over inclined roads 276 
force of wind per sq. ft. for various 

velocities 277 

grade data 292 
grades, effects of, upon load horse can 

draw on different pavements 274 
gross load horse can draw on different 

kinds of road surfaces 274 
impact coefficient, values of 102 
impact stresses in a Pratt truss 103 
lacing bars, thickness of 229 
loads, equivalent uniform 24 
masonry bearings, length of 158 
maximum moments in a deck plate- 
girder 126 
maximum moments, determination of 

position of wheel loads for 93 
maximum shears in a deck plate- 
girder 133 
maximum shear, determination of 

position of wheel loads for 90 
pavements, comparative cost of 

various 398 

pavements, li.'e term of various 394 

pins for country highway bridges 245 

pins for double-track bridges 244 

pins for single-track bridges 244 
plate-girder bridges,, width of, for 

various spans 145 

reactions for a deck plate-girder 126 

reactions for a through plate-girder 123 
resistance due to gravity on different 

inclinations 271 
resistance to traction on different 

pavements 392 




resistance to traction on different 

road surfaces 268 

rise, suitable proportions for different 

paving materials 297 

rivet spacing in bottom flange 177 

rivet spacing in top flange 179 

safe spans for I-beams 151 

specific gravity, weight, resistance to 
crushing, and absorption 
power of stones 352 

standard gauges for angles 175 

stresses in a Pratt truss 1 00 

tension members 217 

tractive power of horses at different 

velocities 273 

Trinidad asphaltum, average com- 
position of 377 
trusses, chronological list of 16 
wheel position, moments in deck 

plate-girder 1 25 

wheel position, moments in a through 

plate-girder 122 

wheel position, shears in a through 

plate-girder 129 

wind stresses in Pratt truss 113 

Through bridges, definition of 14 

Through Pratt railway-span design 203 

determination of span 204 

end-post 239 

floor-beams 207 

intermediate posts 225 

lateral systems 251 

masonry plan 203 

pins 243 

portal 245 

shoes and roller nests 254 

Note. For page numbers see foot of pages. 

Through Pratt railway-span design 


tension members 

top chord 

transverse bracing 
Transverse grade 
Trinidad asphaltum 
Truss bridge development 
Truss, members of 
Truss bridge, definition of 
Trusses, classes of 

chord characteristics 

web characteristics 
Trusses, definition of 
Trusses under dead and live loads 





Trusses under engine loads 


Warren truss 

under dead loads 

under live load 
Water breaks 
Web splice 
Wind-load stresses 

bottom lateral bracing 

overturning effect of wind on train 

overturning effect of wind on truss 

portals and sway bracing 

top lateral system 
Wind loads 
W r ood pavemenis 

























Los Angeles 
This book is DUE on the last date stamped below. 

JUL7E 1 

Form L9-100m-9,'52(A3105)444 



000688173 4